{"id":262,"date":"2024-04-01T14:04:48","date_gmt":"2024-04-01T13:04:48","guid":{"rendered":"https:\/\/tutors4you.com\/?page_id=262"},"modified":"2024-04-01T14:04:49","modified_gmt":"2024-04-01T13:04:49","slug":"binomial-theorem-tutorial","status":"publish","type":"page","link":"https:\/\/tutors4you.com\/index.php\/binomial-theorem-tutorial\/","title":{"rendered":"Binomial Theorem Tutorial"},"content":{"rendered":"\n\n\t\t\n\t\t
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Binomial expression:<\/font>  An algebraic expression consisting of\n two terms with a positive or negative sign between them\n is called a binomial expression.<\/font><\/p>\n

 <\/font>Example:\n (a+b), ( P \/ x2<\/sup>)  – (Q \/ x4<\/sup>)\n etc.<\/font><\/p>\n

 Binomial Theorem:<\/font> When a binomial expression is raised to a\n power ‘n’ we would like to be able to expand\n it. The binomial theorem assists us in doing this. It\n converts such an expression into a series.<\/font><\/p>\n

 Binomial Theorem for positive\n integral index:<\/font><\/p>\n

 (x+y)n<\/sup> = xn<\/sup> +\n n<\/sup>C1<\/sub>xn-1<\/sup>y+n<\/sup>C2<\/sub>xn-2<\/sup>y2<\/sup>+—–+n<\/sup>Crx<\/sub>n-r<\/sup>yr<\/sup>+\n ——-+———+n<\/sup>Cn-1<\/sub>xyn-1  \n <\/sup>+ n<\/sup>cn<\/sub>yn<\/sup>.<\/font><\/p>\n

It can be represented as:<\/font><\/p>\n

(x+y)n<\/sup> = <\/font>n<\/sup>Cr<\/sub>xn-r<\/sup>yr<\/sup><\/font><\/p>\n

Particular – Cases :<\/font><\/p>\n

 (i)     \n Replacing ‘y’ by ‘-y’, we have :<\/font><\/p>\n

(x-y)n<\/sup> = n<\/sup>Co<\/sub>xyo-n<\/sup>C1<\/sub>xn-1<\/sup>y+n<\/sup>C2<\/sub>xn-2<\/sup>y2<\/sup>——-+(-1)r\n n<\/sup>Cr<\/sub>xn-r<\/sup>yr<\/sup>+——+(-1)n<\/sup>\n n<\/sup>Cn<\/sub>xo<\/sup>yn<\/sup>.<\/font><\/p>\n

 It can be represented as :<\/font><\/p>\n

(x+y)n<\/sup> = (-1)r<\/sup> n<\/sup>Cr<\/sub>xn-r<\/sup>yr<\/sup><\/font><\/p>\n

 (ii)     \n Replacing ‘x’ by ‘1’ and\n ‘y’ by ‘x’, we have :<\/font><\/p>\n

(1+x)n<\/sup> = n<\/sup>Co<\/sub>xo<\/sup>+n<\/sup>C1<\/sub>x+n<\/sup>C2<\/sub>x2<\/sup>+———+n<\/sup>Cr<\/sub>xr<\/sup>+——+n<\/sup>Cn-1<\/sub>xn-1<\/sup>+n<\/sup>Cn<\/sub>xn<\/sup>.<\/font><\/p>\n

                  \n     <\/font><\/p>\n

or       \n = n<\/sup>Cr<\/sub>xr<\/sup><\/font><\/p>\n

 (ii)     \n Replacing ‘x’ by ‘-x’, we have :<\/font><\/p>\n

(1+x)n<\/sup> = n<\/sup>Co<\/sub>xo<\/sup>–n<\/sup>C1<\/sub>x1<\/sup>+n<\/sup>C2<\/sub>x2<\/sup>\n – ———+(-1)r n<\/sup>Cr<\/sub>xr<\/sup>+——+n<\/sup>Cn-1<\/sub>(-1)\n n-1             \n <\/sup>+(-1)n<\/sup>     n<\/sup>Cn<\/sub>xn<\/sup>.<\/font><\/p>\n

 or     = (-1)rn<\/sup>Cr<\/sub>xr<\/sup><\/font><\/p>\n

Properties of Binomial – Expansion\n (x+y)n<\/sup> :<\/font><\/p>\n

(i)        There\n are (n+1) terms in the expansion.<\/font><\/p>\n

(ii)        In\n each term, sum of the indices of ‘x’ and\n ‘y’ is equal to ‘n’.<\/font><\/p>\n

(iii)        In\n any term, the lower suffix of ‘c’ is equal to\n the index of ‘y’, and the index of x = n-(lower\n suffix of c).<\/font><\/p>\n

(iv)            \n Because  n<\/sup>Cr<\/sub> = n<\/sup>Cn-r<\/sub>,\n <\/font><\/p>\n

    so we have :<\/font><\/p>\n

n<\/sup>Co<\/sub> = n<\/sup>Cn<\/sub><\/font><\/p>\n

n<\/sup>C1<\/sub>=n<\/sup>Cn-1<\/sub><\/font><\/p>\n

n<\/sup>C2<\/sub>=n<\/sup>Cn-2<\/sub>\n etc.<\/font><\/p>\n

It follows that the coefficients of terms\n equidistant from the beginning and the ends are equal.<\/font><\/p>\n

 EXAMPLES<\/font><\/p>\n

 (1)    \n Simplify (x+v(x2<\/sup>-1)) + (x- v(x2<\/sup>-1))6<\/sup><\/font><\/p>\n

 Solution: let vx2<\/sup>-1  \n = a, so we have:<\/font><\/p>\n

    \n       (x=a)6<\/sup> + (x-a)6<\/sup><\/font><\/p>\n

 = [x6<\/sup>+6<\/sup>C1<\/sub>x5<\/sup>.a+6<\/sup>C2<\/sub>.x4<\/sup>.a2<\/sup>\n + 6<\/sup>C3<\/sub>x3<\/sup>a3<\/sup> + 6<\/sup>C4<\/sub>x2<\/sup>a4<\/sup>\n + 6<\/sup>C5<\/sub>xa5<\/sup> +6<\/sup>C6<\/sub>a6<\/sup>]<\/font><\/p>\n

 + [x6<\/sup>–6<\/sup>C1<\/sub>x5<\/sup>a+6<\/sup>C2<\/sub>.x4<\/sup>.a2<\/sup>\n – 6<\/sup>C3<\/sub>x3<\/sup>a3<\/sup>\n + 6<\/sup>C4<\/sub>x2<\/sup>a4<\/sup>\n – 6<\/sup>C5<\/sub>xa5<\/sup> +6<\/sup>C6<\/sub>a6<\/sup>]<\/font><\/p>\n

= 2[x6<\/sup>+6<\/sup>C2<\/sub>x4<\/sup>a2<\/sup>+6<\/sup>C4<\/sub>x2<\/sup>a4<\/sup>+6<\/sup>C6<\/sub>a6<\/sup>]<\/font><\/p>\n

= 2[x6<\/sup>+15x4<\/sup>(x2<\/sup>-1)+15x2<\/sup>(x2<\/sup>-1)2<\/sup>+(x2<\/sup>-1)3<\/sup>]<\/font><\/p>\n

= 2[x6<\/sup>+15x6<\/sup>-15x4<\/sup>+15x6<\/sup>+15x2<\/sup>-30x4<\/sup>+x6<\/sup>-1-3x4<\/sup>+3x3<\/sup>]<\/font><\/p>\n

= 2[32x6<\/sup>-48x4<\/sup>+18x2<\/sup>-1]<\/font><\/p>\n

Q2:      In the\n expansion of (x+a)n<\/sup>, if the sum of odd-terms\n be ‘P’ and sum of even be ‘Q’ Prove\n that:<\/font><\/p>\n

(i)      P2<\/sup>-Q2<\/sup>\n = (x2<\/sup>-a2<\/sup>)n<\/sup><\/font><\/p>\n

(ii)      4PQ\n = (x+a)2n<\/sup> – (x-a)2n<\/sup><\/font><\/p>\n

Sol.:    (x+a)n<\/sup>\n = xn<\/sup>+n<\/sup>C1<\/sub>xn-1<\/sup>a\n + n<\/sup>C2<\/sub>xn-2<\/sup>a2<\/sup>+n<\/sup>C3<\/sub>xn-3<\/sup>a3<\/sup>\n + ——- + n<\/sup>Cn<\/sub>an<\/sup><\/font><\/p>\n

                       \n = (xn<\/sup>+n<\/sup>C2<\/sub>xn-2<\/sup>a2<\/sup>\n + ———) + (n<\/sup>C1<\/sub>xn-1<\/sup>a+n<\/sup>C3<\/sub>xn-3<\/sup>a3<\/sup>+\n ——)<\/font><\/p>\n

           \n (x+a)n<\/sup> = P+Q ————————> (1)<\/font><\/p>\n

and (x-a)n<\/sup>      \n = xn <\/sup>–  n<\/sup>C1<\/sub>xn-1<\/sup>a+n<\/sup>C2<\/sub>xn-2<\/sup>a2<\/sup>–n<\/sup>C3<\/sub>xn-3<\/sup>a3<\/sup>\n + —-+ (-1)n n<\/sup>Cn<\/sub>an<\/sup><\/font><\/p>\n

                       \n = (xn<\/sup>+n<\/sup>C2<\/sub>xn-2<\/sup>a2<\/sup>\n + —-) – (n<\/sup>C1<\/sub>xn-1<\/sup>a+n<\/sup>C3<\/sub>xn-3<\/sup>a3<\/sup>+——)<\/font><\/p>\n

           \n (x-a)n<\/sup> = P – Q —————-> (2)<\/font><\/p>\n

Now we have :<\/font><\/p>\n

(1)  P2<\/sup> – Q2<\/sup>   \n = (P+Q) (P-Q)<\/font><\/p>\n

                       \n = (x+a)n<\/sup> (x-a)n<\/sup><\/font><\/p>\n

= P2<\/sup> – Q2<\/sup> = (x2<\/sup>-a2<\/sup>)n<\/sup><\/font><\/p>\n

(2) 4 PQ\n         = (P+Q)2<\/sup>\n – (P-Q)2<\/sup><\/font><\/p>\n

= 4 PQ = (x+a)2n<\/sup> – (x-a)2n<\/sup><\/font><\/p>\n

           \n <\/font><\/p>\n

Q4.      Prove\n that (101)50<\/sup> > (100)50<\/sup> + (99)50<\/sup><\/font><\/p>\n

Sol.:    (101)50<\/sup>           \n = (100+1)50<\/sup><\/font><\/p>\n

                                   \n = (100)50<\/sup> + 50<\/sup>c1<\/sub>(100)49<\/sup>\n + 50<\/sup>c2<\/sub>(100)48<\/sup> +\n ——-+1 ——> (i)<\/font><\/p>\n

           \n (99)50<\/sup>             \n = (100-1)50<\/sup><\/font><\/p>\n

                                   \n = (100)50<\/sup> – 50<\/sup>c1<\/sub>(100)49<\/sup>\n + 50<\/sup>c2<\/sub>(100)48<\/sup> –\n ——–+1 ——>(ii)<\/font><\/p>\n

eq(i) – eq(ii) :<\/font><\/p>\n

(101)50<\/sup> – (99)50<\/sup>         \n = 2[50<\/sup>C1<\/sub>(100)49<\/sup> + 50<\/sup>C3<\/sub>(100)47<\/sup>\n + ——–]<\/font><\/p>\n

                                   \n = 2 x   ( 50!\/ 1! X 49!)   (100)49<\/sup>\n + 2. 50<\/sup>C3<\/sub> (100)47<\/sup> +\n ——-<\/font><\/p>\n

           \n          <\/font><\/p>\n

                                   \n = 100 x (100)49<\/sup> + (A positive number)<\/font><\/p>\n

                                   \n = (100)50<\/sup> + (A positive number)<\/font><\/p>\n

(101)50<\/sup> – (99)50<\/sup>\n > (100)50<\/sup><\/font><\/p>\n

or (101)50<\/sup> > (101)50<\/sup>\n + (99)50<\/sup><\/font><\/p>\n

General Terms :<\/font> (r +1) th term from beginning in <\/font><\/p>\n

      (x+y)n<\/sup>\n is called general – term, and <\/font><\/p>\n

it is denoted by <\/font><\/p>\n

                       \n Tr+1<\/sub> = n<\/sup>Cr<\/sub>xn-r<\/sup>yr<\/sub><\/font><\/p>\n

Explanation: We know<\/font><\/p>\n

(x-y)n<\/sup> = n<\/sup>Co<\/sub>xn<\/sup>yo<\/sup>+n<\/sup>C1<\/sub>xn-1<\/sup>y1<\/sup>+n<\/sup>C2<\/sub>xn-2<\/sup>y2<\/sup>+—-+n<\/sup>Cn<\/sub>xo<\/sup>yo<\/sup><\/font><\/p>\n

Here:<\/font><\/p>\n

First term\n        T1<\/sub> = n<\/sup>Co<\/sub>xn<\/sup>yo<\/sup><\/font><\/p>\n

                       \n T2<\/sub> = n<\/sup>C1<\/sub>xn-1<\/sup>y1<\/sup><\/font><\/p>\n

                       \n T3<\/sub> = n<\/sup>C2<\/sub>xn-2<\/sup>y2<\/sup><\/font><\/p>\n

                       \n ————————<\/font><\/p>\n

                       \n ————————<\/font><\/p>\n

                       \n ————————<\/font><\/p>\n

                       \n Tr<\/sub> = n<\/sup>Cr-1<\/sub> xn-(r-1)\n <\/sup>yr-1<\/sup><\/font><\/p>\n

Putting r = r+1 in this expression, we\n get:<\/font><\/p>\n

General Term:<\/font> \n Tr+1<\/sub> = n<\/sup>Cr <\/sub>xn-r <\/sup>yr<\/sup><\/font><\/p>\n

Note : ‘Tr<\/sub>’ can be\n used as general terms also.<\/font><\/p>\n

Problem based on General Terms<\/u><\/b><\/font><\/p>\n

Type : 1_ Q4 Find the 7th<\/sup> term\n in the expansion of <\/font><\/p>\n

[4x – (1 \/ 2vx)]13<\/sup><\/font><\/p>\n

Sol : T7<\/sub>        \n = T6+1<\/sub> =   13<\/sup>C6<\/sub>(4x)13-6<\/sup>    \n – (1\/2vx) 6<\/sup> <\/font><\/p>\n

                            \n =   13<\/sup>C6<\/sub>.47<\/sup>x7<\/sup>.    \n 1 \/(26<\/sup>.x3<\/sup>)<\/font><\/p>\n

                             \n =   13<\/sup>C6<\/sub>. 28<\/sup>.x4<\/sup><\/font><\/p>\n

                              =   \n 13!\/ (6!x7!)     . 28<\/sup>. x4<\/sup><\/font><\/p>\n

=    T7<\/sub> =\n 439296x4<\/sup>             \n     <\/font><\/p>\n

Type II :\n          \n Find the coefficient of x-7 in the expansion\n of      (ax\n –    1\/ bx2<\/sup>)   11<\/sup> \n                                                                                        \n           <\/font><\/p>\n

Sol.:  General Term , Tr+1<\/sub>\n =  11<\/sup>Cr<\/sub>(ax)11-r<\/sup>       \n –   (1\/ bx2<\/sup>  ) r<\/sup>                                       \n <\/font><\/p>\n

Tr+1<\/sub>   = (-1)r<\/sup>\n 11<\/sup>Cr<\/sub>. (a11-r<\/sup> \/ br<\/sup>) \n x11-3r<\/sup>     \n ————–>\n (i)                                        \n        <\/font><\/p>\n

Putting 11 – 3r = -7<\/font><\/p>\n

Or 3r = 18<\/font><\/p>\n

  r = 6<\/font><\/p>\n

From (i) to      T7<\/sub>\n = (-1)6<\/sup> 11<\/sup>C6<\/sub>. ( a5<\/sup>\n \/ b6<\/sup>)  x-7<\/sup> \n ————–> (i)<\/font><\/p>\n

Hence, the coefficient of x-7<\/sup>\n in  ax- (1 \/ b x2<\/sup>) 11<\/sup> is 11<\/sup>C6<\/sub>a5<\/sup>b-6<\/sup>             \n <\/font><\/p>\n

Type III\n :         \n Find the term independent of ‘x’\n in    [(3 x2 <\/sup>\/ 2) – \n (1\/ 3x) ] 9<\/sup><\/font><\/p>\n

Sol.: General Term, Tr+1<\/sub>       \n = 9<\/sup>Cr<\/sub>  (3 x2<\/sup> \/ 2) 9-r<\/sup>\n – (1\/3x)   r<\/sup><\/font><\/p>\n

= (-1)r<\/sup> 9<\/sup>Cr<\/sub>\n ( 3\/2) 9-r <\/sup>x18-2r<\/sup>   <\/sub>    (1\n \/ 3r<\/sup>. xr<\/sup>  )        \n  <\/sub>                                               \n <\/font><\/p>\n

  Tr+1<\/sub>     \n = (-1)r9 <\/sup>Cr  <\/sub>(39-2r<\/sup>\n \/ 29-r<\/sup>). x18-3r<\/sup> ——->\n (i)                           \n      <\/font><\/p>\n

Putting 18- 3 r = o<\/font><\/p>\n

r  =  6<\/font><\/p>\n

So, from (i), 7th<\/sup> term is\n independent of ‘x’, and its value is:<\/font><\/p>\n

T7<\/sub> = (-1)6<\/sup> . 9<\/sup>C6<\/sub>.  \n (3-3<\/sup> \/ 23<\/sup>) xo<\/sup><\/font><\/p>\n

=\n          9 ! \/(6!\n X 3!)  .   1\/ (33<\/sup> x 23<\/sup>)     \n <\/font><\/p>\n

=         \n T7<\/sub> =   (7\/18) <\/font><\/p>\n

Pth<\/sup> term from end:<\/font><\/p>\n

‘P’th<\/sup> term from end\n in the expansion of (x+y)n<\/sup> is (n-P+2)th<\/sup>\n term from beginning.<\/font><\/p>\n

Ex.: Find the 4th<\/sup> term from\n the end in the expansion of  [ (x3<\/sup>\/2) \n –  (2\/x2<\/sup>) ] 7<\/sup><\/font><\/p>\n

 Sol.:    4th<\/sup>\n term from end = (7-4+2)th or 5th<\/sup> term from\n beginning.<\/font><\/p>\n

T5<\/sub>       \n = T4+1<\/sub> = 7<\/sup>C4<\/sub>  (x3<\/sup>\/2)7-4<\/sup>\n . (-2\/x4<\/sup>) 4<\/sup><\/font><\/p>\n

  = 7<\/sup>C4<\/sub> \n (x3<\/sup> \/2) 3   <\/sup>   (\n -2\/ x2<\/sup>) 4<\/sup><\/font><\/p>\n

  =    7! \/ (4!\n X 3!) .  (x9<\/sup>\/8) . (16\/ x8<\/sup>)<\/font><\/p>\n

  =  (7.6.5 \/\n 3.2.1)    .2x<\/font><\/p>\n

 T5<\/sub>    =\n 70x<\/font><\/p>\n

Hence ‘4’ term, from the end =\n 70x.<\/font><\/p>\n

Middle Terms: It depends upon the value\n of ‘n’.<\/font><\/p>\n

Case -1 : When ‘n’ is even,\n then total number of terms in (x+y)n<\/sup> is odd.\n So there is only one middle term i.e. [(n\/2) + 1] \n th them is the middle\n term.              \n <\/font><\/p>\n

So we find  (Tn+1<\/sub>\/2). \n th term in this case, if ‘n’ is even.<\/font><\/p>\n

Case II : When ‘n’ is odd, then\n total number of terms in (x+y)n<\/sup> is even. So\n there are two middle terms i.e. (n+1) \/2 th and \n (n+3) \/2 th are true middle terms.<\/font><\/p>\n

so we find T(n+1)\/2<\/sub>  th\n and T(n+3)\/2<\/sub>  th in this case if\n ‘n’ is odd.<\/font><\/p>\n

      <\/font><\/p>\n

Ex.: Find the middle – term in the\n expansion of [ 3x –  (x3<\/sup> \/ 6)]9<\/sup><\/font><\/p>\n

Sol.: Here total no. of terms are 10\n (even). So there are true middle-terms <\/font><\/p>\n

i.e  (9+1) \/ 2 th and  (9+3) \/\n 2 th. So we have to find – out ‘T5<\/sub>’\n and ‘T6<\/sub>’.<\/font><\/p>\n

T5<\/sub> \n       =  T4+1<\/sub> \n = 9<\/sup>C4<\/sub>(3x)9-4<\/sup>  (-x3<\/sup>\n \/ 6) 4<\/sup><\/font><\/p>\n

  =    9! \/ (4! X\n 5!)         .35\n <\/sup>x5  <\/sup>( <\/sup>x12<\/sup> \/ 64<\/sup>)<\/font><\/p>\n

 = (9.8.7.6 \/ 4.3.2.1)  \n    35<\/sup> \/ (24<\/sup> x 34<\/sup>) \n x17<\/sup><\/font><\/p>\n

T5<\/sub>      \n =   (189 \/ 8)    x17<\/sup><\/font><\/p>\n

T6<\/sub> \n       =  T5+1<\/sub> \n = 9<\/sup>C5<\/sub>(3x)9-5<\/sup>  (-x3<\/sup>\n \/ 6) 5<\/sup><\/font><\/p>\n

 =    9! \/ (5! X\n 4!)         .34\n <\/sup>x4<\/sup>     <\/sup>(x15<\/sup>\n \/ 65<\/sup>)<\/font><\/p>\n

= -(9.8.7.6 \/ 4.3.2.1)    \n 34<\/sup>(25<\/sup> x 35<\/sup>)  x19<\/sup><\/font><\/p>\n

T6<\/sub>     \n = – ( 21 \/ 16) x19<\/sup><\/font><\/p>\n

Greatest – term in (1+x)n<\/sup>\n : If ‘Tr<\/sub>’ and ‘Tr+1<\/sub>’\n be the ‘r’ th and (r+1)th terms in the <\/font><\/p>\n

Expansion of (1+x)n<\/sup>, then :<\/font><\/p>\n

Tr+1<\/sub>\n                 \n = n<\/sup>Cr<\/sub>(1)n-r<\/sup> xr<\/sup>\n = n<\/sup>Cr <\/sub>xr<\/sup><\/font><\/p>\n

                       \n And Tr<\/sub>            \n = n<\/sup>Cr-1<\/sub>. xr-1<\/sup><\/font><\/p>\n

So:   Tr+1 <\/sub>\/ Tr<\/sub>  \n = (n<\/sup>Cr<\/sub>  xr<\/sup> \/ n<\/sup>Cr-1<\/sub>\n xr-1<\/sup>)    = (n-r+1)\/r  |x|<\/font><\/p>\n

If ‘Tr+1<\/sub> be the greatest\n term, then Tr+1<\/sub> Tr<\/sub><\/font><\/p>\n

Or  Tr+1<\/sub> \/ Tr<\/sub> \n   1<\/font><\/p>\n

       <\/font><\/p>\n

since (n-r+1) \/ r.  |x|  \n >=1,  where ‘r’ is a ‘+’ ve\n integer.<\/font><\/p>\n

This inequality, changes either to the\n form r<=m+f pr r <= m, where ‘m’ is a\n ‘+’ ve integer and ‘f’ is a fraction.\n So we get:<\/font><\/p>\n

r <= m + f\n —————>    (i)<\/font><\/p>\n

or       r\n <=  m\n ——————>     (ii)<\/font><\/p>\n

In case (i), ‘T’m+1<\/sub>\n is the greatest term, and in case (i) ‘T’m<\/sub>\n and ‘T’m+1<\/sub> are the greatest terms,\n and both are equal.<\/font><\/p>\n

Short-cut: First calculate  \n m   =     | x (n+1) \/ (x +\n 1) |<\/font><\/p>\n

                                               \n <\/font><\/p>\n

Case (1) If ‘m’ is an integer,\n then ‘T’m<\/sub> and ‘T’m+1<\/sub>\n are the greatest terms and both are equal.<\/font><\/p>\n

Case (2)  If ‘m’ is not an\n integer, then T[m]+1<\/sub> will be the greatest\n term, where [.] denotes greatest integer function.<\/font><\/p>\n

Ex.:      Find\n numerically the greatest term in the expansion of (2+3x),\n <\/font><\/p>\n

when x = (3 \/ 2)<\/font><\/p>\n

                       \n <\/font><\/p>\n

Sol.: 1 Method\n :         (2+3x)9<\/sup> \n = 29<\/sup> [1+ 3x \/ 2]  9<\/sup><\/font><\/p>\n

                                                           \n <\/font><\/p>\n

In the expansion of  [(1 + 3x) \/ 2] 9<\/sup>,\n we have\n :                                 \n  <\/font><\/p>\n

Tr+1 <\/sub>\/ Tr   <\/sub>=       \n (9-r+1)\/ r    |3x \/ 2|<\/font><\/p>\n

  = ((10 –\n r)\/r)    | (3\/2) x(3\/2) |  3<\/sup>  \n     \n              \n <\/font><\/p>\n

 =  (10 – r) \/\n r   x    9 \/ 4<\/font><\/p>\n

Tr+1 <\/sub>\/ Tr   <\/sub>   \n =   (90- 9r) \/ 4r<\/font><\/p>\n

Putting        \n Tr+1 <\/sub>\/ Tr   <\/sub>>=  1<\/font><\/p>\n

(90-9r) \/ 4r  >=  1<\/font><\/p>\n

  or    90\n >=  13 r<\/font><\/p>\n

 or   r    \n 90 \/ 13   <\/font><\/p>\n

or    r  \n <=   6    +   \n 12 \/ 13<\/font><\/p>\n

T6+1<\/sub>  or ‘T7<\/sub>’\n is the greatest term.<\/font><\/p>\n

‘T7<\/sub>’ \n in   [1    +   (3x \/\n 2)] 9<\/sup><\/font><\/p>\n

T7<\/sub>   \n =    T6+1<\/sub>   \n =   9<\/sup>C6<\/sub>  (3x \/ 6)\n 6<\/sup><\/font><\/p>\n

 =     9! \/ (3!\n X 6!) .[ (3 \/ 2)    x   (3 \/ 2)]\n 6<\/sup><\/font><\/p>\n

= (9 .8.7 \/ 3.2.1)   \n x   (96<\/sup> \/ 46<\/sup>)<\/font><\/p>\n

=  (3  x  7 x 96<\/sup>)\n \/ 45<\/sup>     \n =   (3 x 7 x 312<\/sup>) \/ 210<\/sup><\/font><\/p>\n

=  7.   (313<\/sup> \/\n 210<\/sup>)<\/font><\/p>\n

So greatest term in (2+ 3x)9<\/sup> \n is :<\/font><\/p>\n

=  29<\/sup>.  7.  (313<\/sup>\n \/ 210<\/sup>)<\/font><\/p>\n

=   (7 x 313<\/sup>) \/ 2<\/font><\/p>\n

                                   \n <\/font><\/p>\n

II- Method :   (2+3x)9<\/sup>  \n =    29<\/sup> [(1 + 3x) \/\n 2]    9<\/sup><\/font><\/p>\n

=   29<\/sup> [1 + 9 \/\n 4]  9<\/sup><\/font><\/p>\n

since x  =  3 \/ 2<\/font><\/p>\n

Here  m   = | x (n + 1) \/\n (x + 1)|   =   | 9\/4 (9+1) \/ 9\/4 + 1|<\/font><\/p>\n

  =   90 \/ 13   <\/font><\/p>\n

So greatest term in the expansion is T[m]+1<\/sub> \n =  T3+1<\/sub>  =  T7<\/sub><\/font><\/p>\n

Now the method is same as in method (1)<\/font><\/p>\n

Greatest Coefficient :<\/font> In any binomial expansion middle-term has\n the greatest.<\/font><\/p>\n

Coefficient. So <\/font><\/p>\n

(i)      \n If ‘n’ is even, then greatest –\n coefficient  =  n<\/sup>Cn\/2<\/sub><\/font><\/p>\n

(ii)       If\n ‘n’ is odd, then greatest – coefficients\n are n<\/sup>C(n+1)\/ 2<\/sub>  and  n<\/sup>C \n (n-1)\/2<\/sub><\/font><\/p>\n

      <\/font><\/p>\n

Properties of Binomial coefficients :<\/font><\/p>\n

(1) The sum of binomial coefficient in (1\n + x)n<\/sup>  is 2n<\/sup>.<\/font><\/p>\n

Proof (1 + x)n<\/sup>  \n =  Co<\/sub>+C1<\/sub>x+C2<\/sub>x2<\/sup>\n + —– +  Cn<\/sub>xn<\/sup>———–> \n (i)<\/font><\/p>\n

Putting x  =  1  :<\/font><\/p>\n

2n<\/sup>  =  Co<\/sub>\n + C1<\/sub> + C2<\/sub> + ———– + Cn<\/sub>           \n ———–> (ii)<\/font><\/p>\n

Ex.: Prove that the sum of the\n coefficients in the expression (1+x – 3x2<\/sup>)2163<\/sup>\n <\/font><\/p>\n

is  ‘-1’.<\/font><\/p>\n

Sol.:  Putting x = 1  in (1 + x\n – 3x2<\/sup>)2163<\/sup><\/font><\/p>\n

Some of the coefficients\n      =  (1 + 1 – 3)2163<\/sup><\/font><\/p>\n

                       \n            \n   \n          = \n (-1)2163<\/sup>   =   -1<\/font><\/p>\n

(2)  The sum of the coefficients of\n the odd-terms in (1+x)n<\/sup> is equal to the sum of\n coefficients of the even terms and each is equal to 2n-1<\/sup>.<\/font><\/p>\n

Proof: Putting x = -1, in eg(1) :<\/font><\/p>\n

O  = Co<\/sub> – C1<\/sub>\n + C2<\/sub> – C3<\/sub> + —— + (-1)n<\/sup>Cn<\/sub><\/font><\/p>\n

and  from (ii):  2n<\/sup> \n =  Co<\/sub> + C1<\/sub> + C2<\/sub> +\n ——— + Cn<\/sub><\/font><\/p>\n

Adding these egn<\/sup>:<\/font><\/p>\n

2n<\/sup> = 2 ( Co<\/sub> + C2<\/sub>\n + C4<\/sub> + —————)<\/font><\/p>\n

or  Co<\/sub> + C2<\/sub> +\n C4<\/sub> + ——- = 2n-1<\/sup>             \n ————> (ii)<\/font><\/p>\n

Subtracting these egn<\/sup>:<\/font><\/p>\n

2n<\/sup> = 2 (C1 + C3 + C5 +\n ————–)<\/font><\/p>\n

or  C1<\/sub> + C3<\/sub> +\n C5<\/sub> + ——- = 2n-1<\/sup>              \n ————>  (iv)  <\/font><\/p>\n

From (iii) and (iv) :<\/font><\/p>\n

C0<\/sub> + C2<\/sub> + C4<\/sub>\n + ——- = C1<\/sub> + C3<\/sub> + C5<\/sub>\n + ——-  =  2n-1<\/sup><\/font><\/p>\n

Ex.: Evaluate the sum of the 8<\/sup>C1<\/sub>\n + 8<\/sup>C3<\/sub> + 8<\/sup>C5<\/sub> +\n 8<\/sup>C7<\/sub><\/font><\/p>\n

Sol.: since n<\/sup>C1<\/sub> + n<\/sup>C3<\/sub>\n + n<\/sup>C5<\/sub> + n<\/sup>C7<\/sub> +\n ——– =  2n-1<\/sup><\/font><\/p>\n

Here n = 8<\/font><\/p>\n

8<\/sup>C1<\/sub> + 8<\/sup>C3<\/sub>\n + 8<\/sup>C5<\/sub> + 8<\/sup>C7<\/sub> =\n (28-1<\/sup>) <\/font><\/p>\n

=  27<\/sup><\/font><\/p>\n

= 128<\/font><\/p>\n

( 8<\/sup>C9<\/sub>, 8<\/sup>C11<\/sub>\n etc. are not possible)<\/font><\/p>\n

Some important results:<\/font><\/p>\n

(i)        In\n the expansion of (1+x)n<\/sup>, coefficient of xr<\/sup>\n = n<\/sup>Cr<\/sub><\/font><\/p>\n

(ii)       In\n the expansion of (1-x)n<\/sup>, coefficient of xr<\/sup>\n = (-1)r<\/sup>. n<\/sup>Cr<\/sub><\/font><\/p>\n

(iii)       If\n ‘n’ is a negative integer or fraction, then<\/font><\/p>\n

(1+x)n<\/sup>  =  1  +\n (n \/ 1!) x  +  [ n (n-1)\/ 2!] x2<\/sup> \n +  [n(n-1)(n-2) \/ 3!] x3<\/sup> + ————-<\/font><\/p>\n

+ [n(n-1)(n-2) ———–(n-r+1) \/ r!]xr<\/sup>\n + ————–<\/font><\/p>\n

Here | x | <1, i.e. – 1<x<1\n is necessary for its validity.<\/font><\/p>\n

(iv)             \n In (1+x)n<\/sup>, general – term Tr+1<\/sub>\n = [n(n-1)(n-2) ————-(n-r+1) \/ r!]. x2<\/font><\/p>\n

(v)               \n n<\/sup>Cr<\/sub> + n<\/sup>Cr-1<\/sub> =\n n+1<\/sup>Cr<\/sub><\/font><\/p>\n

(vi)             \n n<\/sup>Cx<\/sub> = n<\/sup>Cy<\/sub> x = y or x +\n y = n<\/font><\/p>\n

(vii)           \n n<\/sup>Cr<\/sub> = n\/ r.  n-1<\/sup>Cr-1<\/sub><\/font><\/p>\n

      <\/font><\/p>\n

Multinomial theorem : (For a\n ‘+’ve integral index):<\/font><\/p>\n

If n<\/span>N, and x1<\/sub>, x2<\/sub>,\n x3<\/sub>, ——–xm<\/sub> <\/span> C, then<\/font><\/p>\n

(x1<\/sub> + x2<\/sub> + x3<\/sub>\n + ———+xm<\/sub>)n<\/sup>  =\n ?     n! \/ (n1<\/sub>!  n2<\/sub>!\n —nm<\/sub>!)  x1<\/sub>n1<\/sup>, x2<\/sub>n2<\/sup>\n ….xm<\/sub>nm<\/sup><\/font><\/p>\n

Where n1<\/sub>, n2<\/sub>, n3<\/sub>\n ——–, nm<\/sub> are non-negative integers,\n satisfying the condition<\/font><\/p>\n

n1<\/sub> + n2<\/sub> +\n ———–+nm<\/sub> = n<\/font><\/p>\n

Note: The coefficient of x1<\/sub>n1<\/sup>.\n x2<\/sub>n2<\/sup>. ———xm<\/sub>nm<\/sup>\n in the expansion of<\/font><\/p>\n

(x1<\/sub> + x2<\/sub> + x3<\/sub>\n + ——————- + xm<\/sub>)n<\/sup> is :<\/font><\/p>\n

= n! \/ (n1<\/sub>! x n2<\/sub>!\n —nm<\/sub>!)<\/font><\/p>\n

So, general-term in (a+b+c+d)n<\/sup>\n =   n! \/ (p! x q! x r! x s!). ap<\/sup>.bq<\/sup>.cr<\/sup>.ds<\/sup>.      \n <\/font><\/p>\n

Where p+q+r+s = n, and p, q, r, s <\/span> W.<\/font><\/p>\n

(2)  Number of terms in (x1<\/sub>\n + x2<\/sub> + x3<\/sub> + ——— + x<\/sup>m)n<\/sup>\n :       n+m-1<\/sup>Cm-1.<\/sub><\/font><\/p>\n

Ex.: Find the number of terms in the\n expansion of (2x – 3y + 4z)100<\/sup><\/font><\/p>\n

Sol.: Number of terms =   \n 100+3-1<\/sup>C3-1<\/sub> = 102<\/sup>C2<\/sub><\/font><\/p>\n

                                               \n =      102 ! \/ (2! X 100!)<\/font><\/p>\n

                                               \n = (102 x 101) \/ (2 x 1) = 5151<\/font><\/p>\n

General term of a multinomial –\n theorem :<\/font><\/p>\n

Tr+1<\/sub>  = \n                        \n n! \/ (n1<\/sub>! x n2<\/sub>! —n<\/sup>m!)  \n x1<\/sub>n1<\/sup>. x2<\/sub>n2<\/sup>\n ———–xm<\/sub>nm<\/sup><\/font><\/p>\n

EXAMPLES<\/u><\/b><\/font><\/p>\n

Q1.      Find\n the coefficient of x3<\/sup> y4<\/sup> z2<\/sup>\n in the expansion of (2x – 3y + 4z)9<\/sup>.<\/font><\/p>\n

Sol. General Term in (2x – 3y + 4z)9<\/sup><\/font><\/p>\n

 =   \n         9! \/ (n1<\/sub>!\n X n2<\/sub>! X n3<\/sub>!). (2x)n1<\/sup>. \n (-3y)n2<\/sup>.  (4z)n3<\/sup><\/font><\/p>\n

 =   \n         9! \/ (n1<\/sub>!\n X n2<\/sub>! X n3<\/sub>!).  2n1 <\/sup>(-3)n2<\/sup>.\n (4)n3<\/sup>. xn1<\/sup>. yn2<\/sup>. zn3<\/sup><\/font><\/p>\n

Putting n1 = 3,  n2 =  4, \n n3  = 2 :<\/font><\/p>\n

 \n =            \n 9! \/ (3! x 4! x 2!).     23 <\/sup>(-3)4<\/sup>.\n (4)2<\/sup>. x3<\/sup> y4<\/sup> z2<\/sup><\/font><\/p>\n

  = [ 9. 8. 7. 6. 5. 4!\/ (3.2.1.\n 4!.2)] x 8 x 81 x 16     x3<\/sup>\n y4<\/sup> z2<\/sup><\/font><\/p>\n

Coefficient  of x3<\/sup> y4<\/sup>\n z2<\/sup>  =  9 x 8 x 7 x 5 x 8 x 81 x 8<\/font><\/p>\n

= 13063600<\/font><\/p>\n

Greatest coefficient in the expansion of\n (x1 + x2 + ——– + xm)n<\/sup> is<\/font><\/p>\n

=    n! \/ (q!) m-r<\/sup>\n ( q+1!) r<\/sup><\/font><\/p>\n

           \n       \n       <\/font><\/p>\n

Where ‘q’ is the quotient and\n ‘r’ is the remainder, when ‘n’ is\n divided by ‘m’.<\/font><\/p>\n

Ex.: Find the greatest coefficient in the\n expansion of (a + b + c + d) 15<\/sup>.<\/font><\/p>\n

Sol.: Here n = 15,  m = 4<\/font><\/p>\n

15\/4 is quotient 3 and remainder\n 3.                     \n <\/font><\/p>\n

since q  =  3 \n and   r  =  3<\/font><\/p>\n

Hence greatest – coefficient\n =        15! \/ [(3!)\n 4-3<\/sup> x (3+1!)3<\/sup>]<\/font><\/p>\n

=        \n 15! \/ [(3!) x (4!)3<\/sup> ]<\/font><\/p>\n

=\n          15! \/\n (3! x 4! x 4! x 4!)<\/font><\/p>\n

Ex.: Find the coefficient of x7<\/sup>\n in the expansion of (1+3x-2x3<\/sup>)10<\/sup>.<\/font><\/p>\n

Sol.:    General term in\n (1+3x-2x3<\/sup>)10<\/sup><\/font><\/p>\n

=        \n 10! \/ (n1<\/sub>! x n2<\/sub>! x n3<\/sub>!).\n (1)n1<\/sup>  (3x)n2<\/sup>   \n (-2x3<\/sup>)n3<\/sup><\/font><\/p>\n

=        \n 10! \/ (n1<\/sub>! x n2<\/sub>! x n3<\/sub>!).  \n 3n2<\/sup>    (-2)n3<\/sup> \n xn2+3n3<\/sup><\/font><\/p>\n

Where n1<\/sub> + n2<\/sub> + n3<\/sub>\n = 10 ————–> (i)<\/font><\/p>\n

For coefficient of x7<\/sup> \n :  n2<\/sub> + 3n3<\/sub> = 7\n ————-> (ii)<\/font><\/p>\n

From (ii), possible non-negative integral\n values of ‘n2<\/sub>’  and ‘n3<\/sub>’\n are :<\/font><\/p>\n

           \n n2<\/sub>  = 7,  n3<\/sub>  =\n o          \n since from (i) :      n1<\/sub>\n = 3<\/font><\/p>\n

           \n n2<\/sub>  = 1,  n3<\/sub> =\n 2           \n since from(i) :       n1<\/sub>\n = 7<\/font><\/p>\n

or        \n n2<\/sub> = 4,   n3<\/sub> =\n 1           \n since from (i):       n1<\/sub>\n = 5<\/font><\/p>\n

So required coefficient of x7<\/sup>\n :          <\/font><\/p>\n

10! \/ (3! x 7! x 0!) . (3)7<\/sup>\n (-2)0<\/sup>      +  \n 10! \/ (7! x 1! x 2!). (3)1<\/sup>(-2)2      <\/sup>+  \n 10! \/(5! x 4! x 1!). 34<\/sup> (-2)1<\/sup><\/font><\/p>\n

(10. 9. 8  7!) \/ (7!.3.2.1).37<\/sup>  +\n (10. 9. 8. 7!) \/ (7! . 2) x 3 x\n 4      \n –     [(10. 9. 8. 7. 6. 5! ) \/ (5!.\n 4.3.2.1.)] .3.  2.<\/font><\/p>\n

=  10 x 9 x 4 x 36<\/sup> + 10 x\n 9 x 4 x 3 x 4 – 10 x 9 x 7 x 6 x 33<\/sup> x 2<\/font><\/p>\n

= 10 x 9 x 4 (36<\/sup> + 12 – 7\n x 34<\/sup>)<\/font><\/p>\n

=  360 x (729 + 12 – 567)<\/font><\/p>\n

= 62640<\/font><\/p>\n

Some tips on the solution of binomial\n – coefficients:<\/font><\/p>\n

(1) If the difference of the lower\n suffixes of binomial coefficients in each term is same.<\/font><\/p>\n

For  Ex.:  C1<\/sub> C3<\/sub>\n + C2<\/sub> C4<\/sub> + C3<\/sub> C5<\/sub>\n + —— etc.<\/font><\/p>\n

Then :<\/font><\/p>\n

Case -1 : If each term is positive, then<\/font><\/p>\n

(1+x)n<\/sup>  =  C0<\/sub>\n + C1<\/sub>x + C2<\/sub>x2<\/sup> +\n ———— Cn <\/sub>xn<\/sup>\n —————–> (i)<\/font><\/p>\n

Interchanging ‘1’ and\n ‘x’:<\/font><\/p>\n

 (x+1)n<\/sup> = C0<\/sub> xn<\/sup>\n + C1 <\/sub>xn-1<\/sup> + C2<\/sub> xn-2<\/sup>\n + ——— + Cn<\/sub> ————->(ii)<\/font><\/p>\n

Then multiplying (i) and (ii), and equate\n the coefficient to suitable power of ‘x’ on\n both sides.<\/font><\/p>\n

Case –II : If terms are alternately\n positive and negative<\/font><\/p>\n

Then:<\/font><\/p>\n

(1-x)n<\/sup>   = C0<\/sub>\n – C1<\/sub> x + C2<\/sub> x2<\/sup> –\n ————– + (-1)n<\/sup>  Cn <\/sub>xn<\/sup>\n —————> (1)<\/font><\/p>\n

and      (x+1)n<\/sup> \n = C0<\/sub> xn<\/sup> + C1<\/sub> xn-1<\/sup>\n + C2<\/sub> xn-2<\/sup> + ———- + Cn<\/sub>\n ————-> (2)<\/font><\/p>\n

The multiplying (1) and (2), and equate\n the coefficient of suitable power of ‘x’ on\n both sides.<\/font><\/p>\n

Note : [ (Odd – number) \/ 2] = \n 8<\/font><\/p>\n

 (2)     \n If the sum of the lower suffixes of binomial –\n coefficients in each term is same.<\/font><\/p>\n

For Ex.: C0<\/sub> Cn<\/sub> + C1<\/sub>\n Cn-1<\/sub> + C2<\/sub>. Cn-2<\/sub> +\n ——- + Cn<\/sub> C0<\/sub><\/font><\/p>\n

Then:<\/font><\/p>\n

Case – 1 : If each term is positive,\n then<\/font><\/p>\n

(1+x)n<\/sup>  = C0<\/sub> +\n C1<\/sub> x + C2<\/sub> x2<\/sup> –\n ————– + Cn <\/sub>xn<\/sup>\n —————> (1)<\/font><\/p>\n

and      (1+x)n<\/sup> \n = C0<\/sub> xn<\/sup> + C1<\/sub> x + C2<\/sub>\n x2<\/sup> + ———- + Cn<\/sub> xn<\/sup>————->\n (2)<\/font><\/p>\n

Then multiplying (i) and (ii), and equate\n the coefficient of suitable power of ‘x’ on\n both sides.<\/font><\/p>\n

Case –II : If terms are alternately\n positive and negative,<\/font><\/p>\n

The      (1+x)n<\/sup> \n = C0<\/sub> + C1<\/sub> x + C2<\/sub> x2<\/sup>\n – ————– + Cn <\/sub>xn<\/sup>\n —————> (1)<\/font><\/p>\n

and      (1-x)n<\/sup>  \n = C0<\/sub> – C1<\/sub> x + C2<\/sub> x2<\/sup>\n + ———- + (-1)n<\/sup> Cn<\/sub> xn<\/sup>————->\n (2)<\/font><\/p>\n

Then multiplying (i) and (ii) and\n equating the coefficient of suitable power of\n ‘x’ on both side.<\/font><\/p>\n

PROBLEMS<\/u><\/b><\/font><\/p>\n

(1)       Show\n that the middle – term in the expansion of (1+x)2n<\/sup>\n is<\/font><\/p>\n

 \n                      \n 1. 3. 5 ——- (2n-1) \/ (n!)  . 2n <\/sup>xn<\/sup>,\n ‘n’ being a positive integer.<\/font><\/p>\n

Sol.:    The no. of terms in\n (1+x)2n<\/sup>               \n =         \n 2n+1 (odd).<\/font><\/p>\n

           \n It’s ,middle-term = (2n  + 1) \/\n 2        \n =         \n (n+1)th term.<\/font><\/p>\n

Tn+1<\/sub>           \n =    2n<\/sup>Cn<\/sub> xn<\/sup><\/font><\/p>\n

                       \n =    2n! \/ (n! x n!).  xn<\/sup><\/font><\/p>\n

  \n         \n            =  \n 2n (2n-1) —— 4.3.2.1 \/ (n! x n!).  xn<\/sup><\/font><\/p>\n

                       \n =  [{(2n-1) (2n-3)  —– 3.1.}  { 2n\n (2n-2) —— 4.2.}] \/ (n! x n!). xn<\/sup><\/font><\/p>\n

                       \n =  [{1. 3. 5. —- (2n-1)} 2n<\/sup> {1.2 —-\n n}] \/ (n! x n!) . xn<\/sup><\/font><\/p>\n

                       \n =  [{1.3.5—-(2n-1)}. 2n<\/sup>] \/ (n! x\n n!).  xn<\/sup><\/font><\/p>\n

                       \n =  Tn+1<\/sub>    \n =      1. 3. 5 – (2n-1) \/\n (n!).   2n<\/sup> xn<\/sup><\/font><\/p>\n

 (2)     \n Find the term independent of ‘x’ in the\n expansion of <\/font><\/p>\n

 (i)       \n (1+x+2x3<\/sup>)    [(3 x2<\/sup>\n \/ 2) –  (1\/3x)] 9<\/sup><\/font><\/p>\n

           \n <\/font><\/p>\n

(ii)       \n [( x1\/3<\/sup> \/ 2) + x-1\/5<\/sup>]  8<\/sup><\/font><\/p>\n

Sol.:\n (i)           \n (1+x+2x3<\/sup>)  [(3\/2)x2<\/sup> \n –  (1\/3x)] 9<\/sup><\/font><\/p>\n

=         \n (1+x+2x3<\/sup>) { [(3\/2)x2<\/sup>] 9<\/sup>\n –  9<\/sup>C1<\/sub> [(3\/2)x2 <\/sup>]  \n 8<\/sup>  1\/3x  +  ———-\n +        <\/font><\/p>\n

+ 9<\/sup>C6<\/sub> [(3\/2)x2<\/sup>]\n 3<\/sup> (1\/3x)6<\/sup>  – 9<\/sup>C7<\/sub>\n [(3\/2)x2<\/sup>]  2<\/sup> (1\/3x)7<\/sup>\n —}<\/font><\/p>\n

=         \n (1+x+2x3<\/sup>) { [(3\/2)x2 <\/sup>] 9<\/sup>\n – 9<\/sup>C1<\/sub> (37 <\/sup>\/ 28<\/sup>)x15<\/sup>\n + —- +  9<\/sup>C6<\/sub> (1 x 1 \/ 23\n <\/sup>x 33<\/sup>) – 9<\/sup>C7<\/sub>\n 1\/ (22<\/sup> x 35<\/sup>) 1\/ x3<\/sup> \n +\n —-}                                                      \n <\/font><\/p>\n

Term independent of ‘x’ :<\/font><\/p>\n

           \n 9<\/sup>C6<\/sub> x   1\/ (23<\/sup>\n x 33<\/sup>)     \n –   9<\/sup>C7<\/sub>     \n 2 \/ (22<\/sup> x 35<\/sup>)   <\/font><\/p>\n

=         \n     9! \/ (6! x 3!) . 1\/ (8 x\n 27)   –  9!\/(7! x 2!) .   1\/ (2\n x 243)<\/font><\/p>\n

=         \n (9. 8. 7.  6!) \/ ( 6!. 3. 2. 1.) x 1\/(8. 27)\n –   (9. 8. 7!) \/ (7! . 2).  1\/ (2.243)<\/font><\/p>\n

=         \n 7 \/ 18\n            \n –          \n 2 \/ 27\n            \n =         \n 17 \/ 54<\/font><\/p>\n

(ii)       \n   [(1 \/ 2)  x1\/3<\/sup>  + x-1\/5<\/sup>]\n 8<\/sup><\/font><\/p>\n

           \n <\/font><\/p>\n

Sol.:    General Term Tr+1<\/sub>    \n = n<\/sup>Cr<\/sub> [(1\/2) x1\/3<\/sup>] n-r<\/sup>.\n (x-1\/5<\/sup>)  r<\/sup><\/font><\/p>\n

                                                           \n    n-r<\/u><\/sub>     \n -r<\/u><\/sub><\/font><\/p>\n

                                   \n = n<\/sup>Cr<\/sub> [(1\/2) n-r<\/sup>] x 3<\/sup>  \n x  5<\/sup><\/font><\/p>\n

Here       \n n      =  \n 8    \n                  \n <\/font><\/p>\n

                                   \n = 8<\/sup>Cr<\/sub> (1\/2) 8-r<\/sup>  \n x (8-r)\/3<\/sup>    -r\/5<\/sup><\/font><\/p>\n

                                               \n              \n 40 -8r<\/u><\/sub>      <\/font><\/p>\n

           \n Tr+1<\/sub>                 \n = 8<\/sup>Cr<\/sub> (1\/2)  8-r<\/sup> \n x  15<\/sup>   \n —————> (i)<\/font><\/p>\n

                 \n <\/font><\/p>\n

Putting  (40 – 8r) \/\n 15           \n = 0, we have  r = 5<\/font><\/p>\n

From (i), Term independent of ‘x’\n :<\/font><\/p>\n

           \n T6<\/sub>  =  8<\/sup>C5<\/sub>\n (1\/2) 8-5<\/sup><\/font><\/p>\n

           \n =       8! \/ (5! X 3!)\n .  1 \/ 23<\/sup><\/font><\/p>\n

= \n         (8. 7.\n 6.  !5) \/ (5!. 3.2.1) . 1 \/ 8<\/font><\/p>\n

=         \n T6<\/sub>   =   7<\/font><\/p>\n

(3)      \n Find the coefficient of ‘x’ in the expansion of\n (1-2x3<\/sup> + x5<\/sup>) [1 + (1\/x)]8<\/sup><\/font><\/p>\n

Sol.:    (1-2x3<\/sup>\n + 3x5<\/sup>)  [1 + (1\/x)]   8<\/sup><\/font><\/p>\n

                                   \n        <\/font><\/p>\n

 = (1-2x3<\/sup> + 3x5<\/sup>)\n [1 + 8<\/sup>C1<\/sub> (1\/x) + 8<\/sup>C2<\/sub>  \n (1\/ x2<\/sup>) + 8<\/sup>C3<\/sub> (1\/ x3<\/sup>\n ) + 8<\/sup>C4<\/sub> (1\/ x4<\/sup>)+ 8<\/sup>C5<\/sub>\n (1\/ x5<\/sup> )+ — + 8<\/sup>C8<\/sub> (1\/\n x8<\/sup>)<\/font><\/p>\n

coefficient of x \n          = \n -2.  8<\/sup>C2<\/sub> + 3 8<\/sup>C4<\/sub><\/font><\/p>\n

                                   \n = -2.  8! \/ (2! x 6!) +  3.  8! \/ (4! x\n 4!)<\/font><\/p>\n

                       \n = -2.   (8. 7) \/\n 2      + \n 3      (8. 7. 6. 5.) \/ (4.3.2.1)<\/font><\/p>\n

                       \n =  -56 + 210<\/font><\/p>\n

                       \n = 154<\/font><\/p>\n

(4)      \n Prove that the ration of the coefficient of x10<\/sup>\n in (1-x2<\/sup>)10<\/sup> and the term\n independent of ‘x’ in [x – (2\/x)] 10<\/sup>\n is 1 : 32. <\/font><\/p>\n

Sol.:    In (1-x)2<\/sup> \n : Tr+1<\/sub>   =  10<\/sup>Cr<\/sub>\n (-1)r<\/sup> (x2<\/sup>)r<\/sup><\/font><\/p>\n

           \n Putting  r   =  5<\/font><\/p>\n

                       \n T6<\/sub>   =  –10<\/sup>C5<\/sub>\n x10<\/sup><\/font><\/p>\n

Coefficient of x10<\/sup>  = –10<\/sup>C5<\/sub><\/font><\/p>\n

In [x – (2\/x)] : Tr+1<\/sub> \n =  10<\/sup>Cr<\/sub> (-1)r<\/sup> (x)10-r<\/sup>\n (2\/x)r<\/sup><\/font><\/p>\n

           \n = (-1)r<\/sup> 10<\/sup>Cr<\/sub>. 2r<\/sup>.\n x10-2r<\/sup><\/font><\/p>\n

Putting 10 – 2r  = 0<\/font><\/p>\n

           \n  \n r   =   5<\/font><\/p>\n

So term independent of x : T6<\/sub>         \n = (-1)5<\/sup> 10<\/sup>C5<\/sub>. 25<\/sup><\/font><\/p>\n

           \n Hence their\n ratio        = (-10<\/sup>C5<\/sub>)\n : (-32. 10<\/sup>C5<\/sub>)<\/font><\/p>\n

                                               \n =  1  :  32<\/font><\/p>\n

(6)      \n If third term in the expansion of (x + x logx<\/sup>)5<\/sup>\n is 10,00,000. Find the value of ‘x’.<\/font><\/p>\n

Sol.:    Putting log10<\/sub>x<\/sup> \n =  z in the given expression :<\/font><\/p>\n

           \n We have :<\/font><\/p>\n

                                   \n ( x + xz<\/sup>)5<\/sup><\/font><\/p>\n

           \n T3<\/sub>  =  T2+1<\/sub> \n      =  5<\/sup>C2<\/sub>\n (x)5-2<\/sup>   (xz<\/sup>)2<\/sup><\/font><\/p>\n

                                   \n =  5<\/sup>C2<\/sub> x3<\/sup>. x2z<\/sup><\/font><\/p>\n

                                   \n =        5! \/ (2! x\n 3!) x2z+3<\/sup><\/font><\/p>\n

                                   \n <\/font><\/p>\n

                                   \n =     (5 x 4) \/ 2! x2z+3<\/sup><\/font><\/p>\n

                                   \n = T3<\/sub> = 10x2z+3<\/sup><\/font><\/p>\n

                                   \n 10,00,000 =\n 10. x2z+3<\/sup><\/font><\/p>\n

                                   \n Or x2z+3<\/sup>  =  105<\/sup><\/font><\/p>\n

                                   \n (10z<\/sup>)2z+3<\/sup>  \n =  105<\/sup><\/font><\/p>\n

                                   \n or  102z2+3z<\/sup>   =  105<\/sup><\/font><\/p>\n

                                    2z2<\/sup> +\n 3z  =  5<\/font><\/p>\n

                                   \n [Log10<\/sub>x<\/sup>  =   z]<\/font><\/p>\n

                       \n or 2z2<\/sup> + 3z – 5 = 0<\/font><\/p>\n

           \n <\/font><\/p>\n

                       \n or (z-1) (2z+5)  =  0<\/font><\/p>\n

                   z \n =  1,  –  5 \/ 2<\/font><\/p>\n

                                   \n <\/font><\/p>\n

                       \n or log10<\/sub>x<\/sup> =  1  or \n log10<\/sub>x<\/sup>     \n =  –  5 \/ 2<\/font><\/p>\n

                                   \n <\/font><\/p>\n

since\n x  =  10   or  10-5\/2<\/sup><\/font><\/p>\n

 (7)     \n If in the expansion of (1+x)m<\/sup>  (1-x)n<\/sup>, \n the coefficients of ‘x’ and ‘x2<\/sup>’\n are ‘3’ and ‘-6’ res. Find the value\n of ‘m’.<\/font><\/p>\n

Sol.:  (1+x)m<\/sup>  (1-x)n<\/sup>  \n =  [m<\/sup>C0<\/sub> + m<\/sup>C1<\/sub>x\n + m<\/sup>C2<\/sub>x2<\/sup> + —- +  m<\/sup>Cm<\/sub>\n xm<\/sup>]<\/font><\/p>\n

                                                           \n [n<\/sup>Co<\/sub> – n<\/sup>C1<\/sub>x\n + n<\/sup>C2<\/sub>x2<\/sup> + ——-\n +  (-1)n<\/sup> n<\/sup>Cn<\/sub>xn<\/sup>]<\/font><\/p>\n

Coefficient of x  =  m<\/sup>C1<\/sub>\n x n<\/sup>Co<\/sub> – m<\/sup>C0<\/sub>.\n n<\/sup>C1<\/sub><\/font><\/p>\n

                       \n =        m! \/ ( 1! x\n m-1!)   x 1 – 1 x n! \/ (1! x n-1!)<\/font><\/p>\n

                       \n <\/font><\/p>\n

                       \n =  m – n = 3 —————> (i)<\/font><\/p>\n

Coefficient of x2<\/sup> \n        =  –m<\/sup>C1<\/sub>\n x n<\/sup>C1<\/sub> +n<\/sup>C0<\/sub> x m<\/sup>C2<\/sub>\n + m<\/sup>C0<\/sub> x n<\/sup>C2<\/sub><\/font><\/p>\n

                                   \n =   –     m!\/ (1! x m-1!) x\n n! \/ (1! x n-1!)+ 1 x m! \/ (2! x m-2!) \n     + 1 x n! \/ (2! x n-2!)<\/font><\/p>\n

                                   \n = -mn + m (m-1) \/ 2  +    n(n-1) \/\n 2   =  -6<\/font><\/p>\n

or – 2mn + m(m-1) + n(n-1) \n =  -12<\/font><\/p>\n

or -2mn + m2<\/sup> – m + n2<\/sup>\n – n = 12<\/font><\/p>\n

or (m-n)2<\/sup> – (m+n) = -12<\/font><\/p>\n

From  (i), putting the value of\n (m-n) :<\/font><\/p>\n

                       \n – 9 + (m + n) = 12<\/font><\/p>\n

                       \n or m + n  =  21 ———–> (ii)<\/font><\/p>\n

egn<\/sup> (i) + egn<\/sup>(ii) \n =  2m  =  24<\/font><\/p>\n

           \n <\/font><\/p>\n

m  =  12<\/font><\/p>\n

Q8.      If the\n coefficients of (2r + 1)th term and (r + 2)th term in the\n expansion of (1 + x)43<\/sup> are equal, find\n ‘r’.<\/font><\/p>\n

Sol.:  In (1 + x)43<\/sup> \n :  T2r+1<\/sub>  =  43<\/sup>C2r<\/sub>. \n x2r<\/sup><\/font><\/p>\n

                                   \n Coefficient  =  43<\/sup>C2r<\/sub><\/font><\/p>\n

           \n And Tr+2<\/sub>         \n = 43<\/sup>Cr+1<\/sub>  xr+1<\/sup><\/font><\/p>\n

                                   \n Coefficient = 43<\/sup>Cr<\/sub><\/font><\/p>\n

According to the questions:<\/font><\/p>\n

                       \n 43<\/sup>C2r<\/sub>  =  43<\/sup>Cr+1<\/sub><\/font><\/p>\n

                       \n 2r + r + 1 = 43<\/font><\/p>\n

                       \n or 3r = 42<\/font><\/p>\n

                       \n r  =  14<\/font><\/p>\n

           \n <\/font><\/p>\n

Q9.      If the\n coefficient of ‘4’th and ‘13’th terms\n in the expansion of  [x2<\/sup> +  (1\/x)] n<\/sup>\n be equal, then find the term which independent of\n ‘x’.                              \n       <\/font><\/p>\n

Sol.: T4<\/sub>  = T3+1<\/sub>  \n =  n<\/sup>C3<\/sub> (x2<\/sup>)n-3<\/sup>.    \n 1\/ x3<\/sup><\/font><\/p>\n

                                               \n         <\/font><\/p>\n

                       \n Coefficient =  n<\/sup>C3<\/sub><\/font><\/p>\n

           \n T13<\/sub>  =  T12+1<\/sub> \n =  n<\/sup>C12<\/sub> (x2<\/sup>)n-12<\/sup>  \n 1 \/ x12<\/sup><\/font><\/p>\n

                                   \n <\/font><\/p>\n

                       \n Coefficient = n<\/sup>C12<\/sub><\/font><\/p>\n

According to the question:<\/font><\/p>\n

                       \n n<\/sup>C3<\/sub>  =  n<\/sup>C12<\/sub><\/font><\/p>\n

                        n  = \n 12 + 3<\/font><\/p>\n

                       \n n  =  15<\/font><\/p>\n

           \n <\/font><\/p>\n

  Expansion =  [x2<\/sup> \n + (1\/x)]15<\/sup><\/font><\/p>\n

Now Tr+1<\/sub>  =  15<\/sup>Cr<\/sub>.\n (x2<\/sup>)15-r<\/sup>.  1\/ xr<\/sup><\/font><\/p>\n

                                   \n          <\/font><\/p>\n

           \n Tr+1<\/sub>  =  15<\/sup>Cr<\/sub>.\n x30-3r<\/sup>  ————-> (i)<\/font><\/p>\n

Putting :<\/font><\/p>\n

           \n 30 – 3r  =  0<\/font><\/p>\n

           \n  \n r     =  10<\/font><\/p>\n

From (i) T11<\/sub>  =  15<\/sup>C10<\/sub>  \n =      15!\/(10! x 5!)  =\n (15 x 14 x 13 x 12 x 11) \/ (5 x 4 x 3 x 2 x 1)<\/font><\/p>\n

                                   \n =  3003.<\/font><\/p>\n

Q10.    In the expansion\n of (a – b)n<\/sup>, n <= 5, if the sum of the\n 5th<\/sup> and 6th<\/sup> terms is zero. Find ( a\n \/ b)  in terms of ‘n’.<\/font><\/p>\n

Sol.:  T5<\/sub>  =  T4+1<\/sub>  \n =  n<\/sup>C4<\/sub> (a)n-4<\/sup> (-b)4<\/sup><\/font><\/p>\n

                       \n T5<\/sub>  =  n<\/sup>C4<\/sub> an-4<\/sup>\n b4<\/sup><\/font><\/p>\n

                       \n T6<\/sub>  =  T5+1<\/sub>  = \n n<\/sup>C5 <\/sub>(a)n-5<\/sup> (-b)5<\/sup> \n =  –n<\/sup>C5<\/sub> an-5<\/sup> b5<\/sup><\/font><\/p>\n

                       \n T5<\/sub> + T6<\/sub>  =  0<\/font><\/p>\n

                       \n n<\/sup>C4<\/sub>\n an-4<\/sup> b4<\/sup> – n<\/sup>C5<\/sub>\n an-5<\/sup> b5<\/sup>  =  0<\/font><\/p>\n

                       \n or  n<\/sup>C4<\/sub> an-4 <\/sup>b4<\/sup> \n =  n<\/sup>C5<\/sub> an-5<\/sup> b5<\/sup><\/font><\/p>\n

  \n         \n or        \n       n!\/(4! x n-4!) an-4<\/sup> \n =          \n n!\/(5! x n-5!)  an-5<\/sup> b<\/font><\/p>\n

           \n <\/font><\/p>\n

           \n or            \n an-4<\/sup> \/ (n-4)(n-5!)    = \n         \n     an-5<\/sup> \/ 5(n-5!) b<\/font><\/p>\n

           \n or        \n      an-4<\/sup> \/ an-5<\/sup>               \n   =        b \/ 5\n (n-4)<\/font><\/p>\n

           \n or         a(n-4)-(n-5)<\/sup>          \n =          (n\n – 4) \/ 5  .b<\/font><\/p>\n

                                               \n    \n          <\/font><\/p>\n

           \n or        \n a         \n =         \n (n – 4)\/5 . b<\/font><\/p>\n

           \n or     \n a\/b        \n =   (n – 4) \/ 5<\/font><\/p>\n

        \n <\/font><\/p>\n

Q11.    Find the\n coefficient of xr<\/sup>  in the expansion\n of   [x + (1\/x)] n<\/sup>, if it occurs.<\/font><\/p>\n

Sol.:    General term : Tp+1<\/sub> \n =  n<\/sup>Cp<\/sub> (x)n-p<\/sup> \n (1\/x) p<\/sup><\/font><\/p>\n

                                               \n <\/font><\/p>\n

                       \n Tp+1<\/sub>  =  n<\/sup>Cp<\/sub> xn-2p<\/sup> \n —————> (i)<\/font><\/p>\n

           \n Putting  n-2p  =  r<\/font><\/p>\n

                       \n \n p   =      \n (n  –  r) \/ 2<\/font><\/p>\n

From:\n (i)         T (n-r)\n \/ 2<\/sub>  +1<\/sub>   =  n<\/sup>C(n-r)\n \/ 2<\/sub> xr<\/sup><\/font><\/p>\n

                       \n Coefficient of xr<\/sup>  =  n<\/sup>C (n\n – r) \/ 2<\/sub><\/font><\/p>\n

                                   \n <\/font><\/p>\n

Q12.:   Prove that the\n coefficient of the term independent of ‘y’ in\n the expansion of<\/font><\/p>\n

                     \n [(y + 1)\/( y 2\/3<\/sup> – y1\/3<\/sup> + 1)\n –       (y – 1) \/ (y\n – y1\/2<\/sup>)]10<\/sup>  is 210.<\/font><\/p>\n

           \n <\/font><\/p>\n

Sol.:    We\n have        (y + 1) \/\n (y 2\/3<\/sup> – y1\/3<\/sup> + 1)<\/font><\/p>\n

                       \n      <\/font><\/p>\n

           \n Putting y  =  t3<\/sup>, we have<\/font><\/p>\n

=         \n   (t3<\/sup> + 13<\/sup>) \/ (t2<\/sup>\n – t + 1) =       \n (t + 1) (t2<\/sup> – t + 1) \/ (t2<\/sup>\n – t + 1)<\/font><\/p>\n

                                               \n            \n    <\/font><\/p>\n

=         \n t + 1     <\/font><\/p>\n

      (y + 1) \/\n (y2\/3<\/sup> – y1\/3<\/sup> +\n 1)               \n =          y1\/3<\/sup>\n + 1<\/font><\/p>\n

and Putting y  =  a2<\/sup> \n in     (y – 1) \/ (y – y1\/2<\/sup>\n ) :<\/font><\/p>\n

                       \n <\/font><\/p>\n

                       \n =     (a2<\/sup> – 1) \/ (a2<\/sup>\n – a) =   (a+1) (a-1) \/ [a (a-1)]<\/font><\/p>\n

                       \n =  (a + 1) \/ a   =  1  +  1 \/ a<\/font><\/p>\n

    (y – 1) \/ (y – y1\/2<\/sup>)\n =  1 +  1 \/vy <\/font><\/p>\n

     \n      \n            \n <\/font><\/p>\n

    (y + 1) \/ (y2\/3<\/sup>\n – y1\/3<\/sup> +1)  –   (y –\n 1) \/ (y – y1\/2<\/sup>)]10<\/sup>\n =    [y1\/3<\/sup> + 1 – 1\n –   (1\/ y1\/2<\/sup>)]   10<\/sup><\/font><\/p>\n

                                               \n = (y1\/3<\/sup> – y-1\/2<\/sup> )10<\/sup><\/font><\/p>\n

In ( y1\/3<\/sup> – y-1\/2<\/sup>)10<\/sup>,<\/font><\/p>\n

                       \n Tr+1<\/sub>      = 10<\/sup>Cr<\/sub>\n (y1\/3<\/sup>)10-r<\/sup>. (-y-1\/2<\/sup>)r<\/sup><\/font><\/p>\n

                                   \n     <\/font><\/p>\n

           \n =         \n (-1)r<\/sup> 10<\/sup>Cr<\/sub>. <\/sub>(10-r)\n \/ 3   –  r\/2<\/sup> <\/font><\/p>\n

                                   \n        <\/font><\/p>\n

Tr+1<\/sub>     \n =         \n (-1)r<\/sup> 10<\/sup>Cr<\/sub>. y(20-5r)\n \/ 6<\/sup> <\/font><\/p>\n

Putting           \n (20 – 5r) \/ 6 = 0<\/font><\/p>\n

           \n or     r = 4<\/font><\/p>\n

Putting this value in (1)  T5<\/sub> \n =  (-1)4<\/sup>  10<\/sup>C4<\/sub><\/font><\/p>\n

                                               \n =    10!\/ (6! x 4!)   \n =    (10 x 9 x 8 x 7) \/ (4 x 3 x 2 x 1)<\/font><\/p>\n

                                   \n T5<\/sub>       \n =  210<\/font><\/p>\n

Q13:    x4-r<\/sup>\n occurs in the expansion of [x + (1\/ x2<\/sup>)] 4n<\/sup>,\n prove that its coefficients is:<\/font><\/p>\n

           \n <\/font><\/p>\n

                       \n = (4n!) \/[ (4\/3)n-r]! x [(4\/3)(2n+r)]!<\/font><\/p>\n

Sol.:    In   [x\n + (1\/x2<\/sup>)]4n<\/sup>, Tp+1<\/sub>  \n =  4n<\/sup>Cp  <\/sub>(x)4n-p<\/sup>\n (1\/ x2<\/sup>)p<\/sup><\/font><\/p>\n

           \n Tp+1<\/sub>  =  4n<\/sup>Cp<\/sub> \n x4n-3p<\/sup>  ———-> (i)<\/font><\/p>\n

Putting :<\/font><\/p>\n

                       \n 4n – 3p    =  4r<\/font><\/p>\n

           \n or         4 (\n n-r ) \/ 3     = p<\/font><\/p>\n

From (i) \n         Tp+1<\/sub>    \n =   4n<\/sup>C4(n-r) \/ 4<\/sub>. x4r<\/sup><\/font><\/p>\n

Coefficient of x4r<\/sup> \n       =  4n<\/sup>C4<\/sub>\n (n-r) \/ 3<\/sub><\/font><\/p>\n

                                   \n = (4n!) \/ [(4\/3)n-r]! x [(4n\/1) – 4(n-r)\/3]!<\/font><\/p>\n

                                   \n = (4n!) \/ [(4\/3)n-r]! x [(4\/3) 2(n+r)]!   \n <\/font><\/p>\n

           \n <\/font><\/p>\n

Q14.    Find the\n coefficient of x50<\/sup> in (1+x)41<\/sup>\n (1-x+x2<\/sup>)40<\/sup>.<\/font><\/p>\n

Sol.:    (1+x)41<\/sup>\n (1-x+x2<\/sup>)40<\/sup>      \n = (1+x) (1+x)40<\/sup> (1-x+x2<\/sup>)40<\/sup><\/font><\/p>\n

                                               \n = (1+x) [ (1+x) ( 1-x + x2<\/sup>)]40<\/sup><\/font><\/p>\n

                                               \n = (1+x) (1+x3<\/sup>)40<\/sup><\/font><\/p>\n

General Term = Tr+1<\/sub>             \n = (1+x) [40<\/sup>Cr<\/sub> (x3<\/sup>)r<\/sup>]<\/font><\/p>\n

                                               \n = 40<\/sup>Cr<\/sub> (1+x) x3r<\/sup><\/font><\/p>\n

                                               \n = 40<\/sup>Cr<\/sub> (x3r<\/sup> + x3r +\n 1<\/sup>)<\/font><\/p>\n

Here either 3r = 50  or 3r+1 \n =  50<\/font><\/p>\n

          \n r   =  (50 \/ 3)  or  (49 \/ 3)<\/font><\/p>\n

The value of ‘r’ is a fraction,\n so it doesn’t contain the term x50<\/sup>. So\n coefficient of x50<\/sup> is ‘0’.<\/font><\/p>\n

Q15.:   Show that that the term\n independent of ‘x’ in the expansion of <\/font><\/p>\n

           \n [x  +  (1\/x)]  2n<\/sup> \n is     [1. 3. 5.  —- (2n-1) \/\n (n!)] 2n<\/sup><\/font><\/p>\n

                                   \n <\/font><\/p>\n

Sol.:    General Term Tr+1<\/sub>    \n = 2n<\/sup>Cr <\/sub>(x)2n-r<\/sup> \n (1\/x) r<\/sup><\/font><\/p>\n

                                               \n = 2n<\/sup>Cr<\/sub>.  x2n-2r<\/sup>\n ———> (i)<\/font><\/p>\n

           \n Here 2n – 2r  =  0<\/font><\/p>\n

                       \n or    n  =  r<\/font><\/p>\n

From\n (i)         \n Tr+1    = 2n<\/sup>Cn<\/sub><\/font><\/p>\n

                                   \n =       2n! \/ (n! x n!)<\/font><\/p>\n

                                   \n = [2n (2n-1 ) —— 3. 2. 1] \/  ( n! x n!)<\/font><\/p>\n

                                   \n = { 2n (2n-2)  —- 4. 2 } { (2n-1) (2n-3) —–\n 3.1.} \/ (n! x n!)<\/font><\/p>\n

                                   \n = [2n<\/sup> {n (n-1)  —–2.1.}]  {\n (2n-1) ——- 4.3.1.} \/ (n! x n!)<\/font><\/p>\n

                                   \n = 2n.  n!{(2n-1) —- 5. 3. 1. \/ (n! x n!)<\/font><\/p>\n

                                   \n <\/font><\/p>\n

                                   \n = {1. 3. 5. —– (2n -1)} 2n<\/sup> \/ n!<\/font><\/p>\n

                                               \n     <\/font><\/p>\n

Q16.    The 3rd<\/sup>,\n 4th<\/sup> and 5th<\/sup> terms in the expansion\n of (x+a)n<\/sup> are respectively ‘84’,\n ‘280’ and ‘560’, find the value of\n ‘x’, ‘a’ and ‘n’.<\/font><\/p>\n

Sol.:    Tr+1<\/sub> \n = n<\/sup>Cr<\/sub>  xn-r<\/sup>. ar<\/sup><\/font><\/p>\n

           \n Putting r  =  2, 3 and 4 respectively<\/font><\/p>\n

           \n T3<\/sub> = n<\/sup>C2<\/sub> xn-2<\/sup>.\n a2<\/sup>\n          = 84\n ————>(i)<\/font><\/p>\n

           \n T4<\/sub> = n<\/sup>C3<\/sub> xn-3<\/sup>\n a3<\/sup>          \n = 280———->(ii)<\/font><\/p>\n

and      T5<\/sub>\n = n<\/sup>C4<\/sub> xn-4<\/sup> a4<\/sup>          \n = 560 ——–>(iii)<\/font><\/p>\n

egn (i) x egn(iii) : [n<\/sup>C2<\/sub>\n xn-2<\/sup> a2<\/sup>] [n<\/sup>C4<\/sub>\n xn-4<\/sup> a4<\/sup>]  \n    =  84 x 560<\/font><\/p>\n

                       \n =       n!\/[2! x\n (n-2)!]   x  n! \/ [4! x (n-4)!] . x2n-6<\/sup>\n a6<\/sup>  =  84 x 560<\/font><\/p>\n

                       \n <\/font><\/p>\n

           \n or n (n-1) \/ 2   x  n(n-1) (n-2) (n-3) \/\n 4! x   x2n-6<\/sup> a6<\/sup>   \n =  84 x 560 ——-> (iv)<\/font><\/p>\n

Squaring of egn<\/sup> (ii), we have\n :<\/font><\/p>\n

           \n (n<\/sup>C3<\/sub>  x n-3<\/sup> a3<\/sup>)2<\/sup>   \n =  2802<\/sup><\/font><\/p>\n

     n<\/sup>C3<\/sub> \n x n<\/sup>C3<\/sub>  x x2n-6<\/sup> \n x  a6<\/sup>      \n =  2802<\/sup><\/font><\/p>\n

           \n =         \n          n!\n \/ [3! x (n-3)!] x n! \/ [3! x (n-3)!] x  x2n-6<\/sup>\n a6<\/sup>  =  2802<\/sup><\/font><\/p>\n

           \n or n (n-1)(n-2) \/ 6   x  n(n-1) (n-2)\n (n-3) \/ 3! x   x2n-6<\/sup> a6<\/sup>   \n =  280 x 280 ——-> (v)<\/font><\/p>\n

egn (v) egn(iv) :<\/font><\/p>\n

n2<\/sup> (n-1)2<\/sup> (n-2)2 <\/sup>\/\n (6 x 3!) x  2 x  4! \/ [n2<\/sup>(n-1)2<\/sup>\n (n-2)(n-3)] = (280 x 280) \/ (84 x 560) <\/font><\/p>\n

or        \n 4 (n-2) \/ 3 (n-3)\n =        5 \/ 3<\/font><\/p>\n

or        \n 4n – 8  =  5n – 15<\/font><\/p>\n

n \n =  7<\/font><\/p>\n

Putting this value in (i), (ii) and (iii)\n :<\/font><\/p>\n

           \n 7<\/sup>C2<\/sub> x5<\/sup> a2<\/sup>        \n =         \n    84 —————> (vi)<\/font><\/p>\n

           \n 7<\/sup>C3<\/sub> x4<\/sup> a3<\/sup>        \n =         \n 280 —————->(vii)<\/font><\/p>\n

           \n 7<\/sup>C4<\/sub> x3<\/sup> a4<\/sup>        \n =         \n 560—————–>(viii)<\/font><\/p>\n

egn<\/sup> (vii) egn<\/sup>(vi):<\/font><\/p>\n

           \n (7<\/sup>C3<\/sub> x4<\/sup> a3<\/sup>)\n <\/sup>\/ (7<\/sup>C2<\/sub> x5<\/sup> a2<\/sup>)\n         \n =         \n 280 \/ 84<\/font><\/p>\n

           \n <\/font><\/p>\n

[7! \/ (3! x 4!)a] \/ [7! \/ (2! x 5!)x] =\n 10 \/\n 3                                \n <\/font><\/p>\n

           \n or                    \n 7! \/ (3! x 4!)   x (2! x 5!) \/ 7! x  a \/\n x     =  \n    10 \/ 3<\/font><\/p>\n

                       \n <\/font><\/p>\n

           \n or                    \n 5 \/ 3     x  a \/ x\n =         \n 10 \/ 3<\/font><\/p>\n

or\n         a  \n        =   2x<\/font><\/p>\n

           \n <\/font><\/p>\n

Putting this\n value in egn (vi):<\/font><\/p>\n

           \n 7<\/sup>C2<\/sub>. x5<\/sup>. 4x2<\/sup>  \n = 84<\/font><\/p>\n

or        \n 7! \/ (2! x 5!)x7<\/sup> \n =         \n 21<\/font><\/p>\n

       (7 \n x  63) \/ 2 x7<\/sup>           \n =         \n 21<\/font><\/p>\n

x7<\/sup>  \n =     1<\/font><\/p>\n

x  =  1<\/font><\/p>\n

Putting this\n value in (ix)  =   a  =  2<\/font><\/p>\n

Q17. Let ‘n’ be a positive\n integer. If the coefficients of second, third and fourth\n terms in (1+x)2<\/sup> are in arithmetic progression,\n then find the value of ‘n’.<\/font><\/p>\n

Sol:     General Term\n : Tr+1<\/sub> = n<\/sup>Cr<\/sub>  xr<\/sup><\/font><\/p>\n

           \n 2nd<\/sup>\n Term : T2<\/sub> = n<\/sup>C1<\/sub> x<\/font><\/p>\n

                       \n Coefficient  =  n<\/sup>C1<\/sub><\/font><\/p>\n

           \n 3rd<\/sup>\n Term :  T3<\/sub> = n<\/sup>C2<\/sub> x2<\/sup><\/font><\/p>\n

                       \n Coefficient  =  n<\/sup>C2<\/sub><\/font><\/p>\n

           \n <\/font><\/p>\n

           \n Similarly coefficient of 4th<\/sup> term  = nC3<\/font><\/p>\n

           \n These are in A. P., so.<\/font><\/p>\n

           \n 2 n<\/sup>C2<\/sub> = n<\/sup>C1<\/sub>  \n +  n<\/sup>C3<\/sub><\/font><\/p>\n

     2 [n! \/ {2! x\n (n-2)!}]       \n    \n =        n! \/ {1! x\n (n-1!)} + n! \/ {3! x (n-3!)}<\/font><\/p>\n

or        \n n! \/\n (n-2!)        \n =     n! [1 \/ (n-1!)  + 1 \/ {6\n (n-3!}]<\/font><\/p>\n

or        \n 1\/ [(n-3) x (n-3!)   = 1 \/ [(n-1)x (n-2) x(n\n – 3!)]  +1\/ [6! (n-3!)] )<\/font><\/p>\n

or        \n 1 \/ (n – 2) – 1 \/ [(n-1) (n-2)]   \n =          1\n \/ 6<\/font><\/p>\n

      \n     \n                        \n <\/font><\/p>\n

or        \n (n – 1 – 1) \/ [(n-1)\n (n-2)]        \n =          1\n \/ 6 <\/font><\/p>\n

                                   \n <\/font><\/p>\n

or          \n (n- 2) \/ [(n-1) (n-2)]\n =          \n 1 \/ 6<\/font><\/p>\n

or        \n n – 1     =   6<\/font><\/p>\n

n  =  7<\/font><\/p>\n

Q18.    The 6th<\/sup>\n term in the expansion of [(1\/ x8\/3<\/sup>) + x2<\/sup>\n log10<\/sub>x<\/sup>]8<\/sup> is 5600. Prove\n that x =10.  <\/font><\/p>\n

Sol.:    T6<\/sub> = T5+1<\/sub>        \n =          8<\/sup>C5<\/sub>\n (1\/ x8\/3<\/sup>) 8-5  <\/sup>(  <\/sup>x2<\/sup>\n log10<\/sub>x <\/sup>)  5<\/sup><\/font><\/p>\n

                                               \n         <\/font><\/p>\n

or  8<\/sup>C5<\/sub> x (1 \/\n x8<\/sup>)   x     c10<\/sup>\n x  (log10<\/sub>x<\/sup>)5<\/sup>\n =     5600<\/font><\/p>\n

           \n     <\/font><\/p>\n

      8! \/ (5! x 3!)\n x    c2<\/sup> (log10<\/sub>x<\/sup>)5<\/sup>  \n = 5600<\/font><\/p>\n

   8. 7. 6. \/ 6 x   \n c2<\/sup> (log10<\/sub>x<\/sup>)5<\/sup>  \n = 5600<\/font><\/p>\n

or x2<\/sup>(log10<\/sub>x<\/sup>)5<\/sup>  \n = 100 = 102<\/sup><\/font><\/p>\n

Clearly    x = 10\n satisfied as log10<\/sub>10<\/sup>  = \n 1.<\/font><\/p>\n

If  x  > 10  \n or  <  10, the result will change in\n inequality.<\/font><\/p>\n

———–x——————–x————————-x———————–x—————-x—–<\/font><\/p>\t\n\t\t<\/div>\n\t\n","protected":false},"excerpt":{"rendered":"

Binomial expression:  An algebraic expression consisting of two terms with a positive or negative sign between them is called a binomial expression.  Example: (a+b), ( P \/ x2)  – (Q \/ x4) etc.  Binomial Theorem: When a binomial expression is raised to a power ‘n’ we would like to be able to expand it. The […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-262","page","type-page","status-publish","hentry"],"blocksy_meta":[],"yoast_head":"\nBinomial Theorem Tutorial - Tutors 4 You<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/tutors4you.com\/index.php\/binomial-theorem-tutorial\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Binomial Theorem Tutorial - Tutors 4 You\" \/>\n<meta property=\"og:description\" content=\"Binomial expression:  An algebraic expression consisting of two terms with a positive or negative sign between them is called a binomial expression.  Example: (a+b), ( P \/ x2)  – (Q \/ x4) etc.  Binomial Theorem: When a binomial expression is raised to a power ‘n’ we would like to be able to expand it. 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