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 binomial theorem assists us in doing this. It converts such an expression into a series.
Binomial Theorem for positive integral index:
(x+y)n = xn + nC1xn-1y+nC2xn-2y2+—–+nCrxn-ryr+ ——-+———+nCn-1xyn-1 + ncnyn.
It can be represented as:
(x+y)n =
nCrxn-ryr
Particular – Cases :
(i) Replacing ‘y’ by ‘-y’, we have :
(x-y)n = nCoxyo-nC1xn-1y+nC2xn-2y2——-+(-1)r nCrxn-ryr+——+(-1)n nCnxoyn.
It can be represented as :
(x+y)n = (-1)r nCrxn-ryr
(ii) Replacing ‘x’ by ‘1’ and ‘y’ by ‘x’, we have :
(1+x)n = nCoxo+nC1x+nC2x2+———+nCrxr+——+nCn-1xn-1+nCnxn.
or
= nCrxr
(ii) Replacing ‘x’ by ‘-x’, we have :
(1+x)n = nCoxo–nC1x1+nC2x2 – ———+(-1)r nCrxr+——+nCn-1(-1) n-1 +(-1)n nCnxn.
or = (-1)rnCrxr
Properties of Binomial – Expansion (x+y)n :
(i) There are (n+1) terms in the expansion.
(ii) In each term, sum of the indices of ‘x’ and ‘y’ is equal to ‘n’.
(iii) In any term, the lower suffix of ‘c’ is equal to the index of ‘y’, and the index of x = n-(lower suffix of c).
(iv) Because nCr = nCn-r,
so we have :
nCo = nCn
nC1=nCn-1
nC2=nCn-2 etc.
It follows that the coefficients of terms equidistant from the beginning and the ends are equal.
EXAMPLES
(1) Simplify (x+v(x2-1)) + (x- v(x2-1))6
Solution: let vx2-1 = a, so we have:
(x=a)6 + (x-a)6
= [x6+6C1x5.a+6C2.x4.a2 + 6C3x3a3 + 6C4x2a4 + 6C5xa5 +6C6a6]
+ [x6–6C1x5a+6C2.x4.a2 – 6C3x3a3 + 6C4x2a4 – 6C5xa5 +6C6a6]
= 2[x6+6C2x4a2+6C4x2a4+6C6a6]
= 2[x6+15x4(x2-1)+15x2(x2-1)2+(x2-1)3]
= 2[x6+15x6-15x4+15x6+15x2-30x4+x6-1-3x4+3x3]
= 2[32x6-48x4+18x2-1]
Q2: In the expansion of (x+a)n, if the sum of odd-terms be ‘P’ and sum of even be ‘Q’ Prove that:
(i) P2-Q2 = (x2-a2)n
(ii) 4PQ = (x+a)2n – (x-a)2n
Sol.: (x+a)n = xn+nC1xn-1a + nC2xn-2a2+nC3xn-3a3 + ——- + nCnan
= (xn+nC2xn-2a2 + ———) + (nC1xn-1a+nC3xn-3a3+ ——)
(x+a)n = P+Q ————————> (1)
and (x-a)n = xn – nC1xn-1a+nC2xn-2a2–nC3xn-3a3 + —-+ (-1)n nCnan
= (xn+nC2xn-2a2 + —-) – (nC1xn-1a+nC3xn-3a3+——)
(x-a)n = P – Q —————-> (2)
Now we have :
(1) P2 – Q2 = (P+Q) (P-Q)
= (x+a)n (x-a)n
= P2 – Q2 = (x2-a2)n
(2) 4 PQ = (P+Q)2 – (P-Q)2
= 4 PQ = (x+a)2n – (x-a)2n
Q4. Prove that (101)50 > (100)50 + (99)50
Sol.: (101)50 = (100+1)50
= (100)50 + 50c1(100)49 + 50c2(100)48 + ——-+1 ——> (i)
(99)50 = (100-1)50
= (100)50 – 50c1(100)49 + 50c2(100)48 – ——–+1 ——>(ii)
eq(i) – eq(ii) :
(101)50 – (99)50 = 2[50C1(100)49 + 50C3(100)47 + ——–]
= 2 x ( 50!/ 1! X 49!) (100)49 + 2. 50C3 (100)47 + ——-
= 100 x (100)49 + (A positive number)
= (100)50 + (A positive number)
(101)50 – (99)50 > (100)50
or (101)50 > (101)50 + (99)50
General Terms : (r +1) th term from beginning in
(x+y)n is called general – term, and
it is denoted by
Tr+1 = nCrxn-ryr
Explanation: We know
(x-y)n = nCoxnyo+nC1xn-1y1+nC2xn-2y2+—-+nCnxoyo
Here:
First term T1 = nCoxnyo
T2 = nC1xn-1y1
T3 = nC2xn-2y2
————————
————————
————————
Tr = nCr-1 xn-(r-1) yr-1
Putting r = r+1 in this expression, we get:
General Term: Tr+1 = nCr xn-r yr
Note : ‘Tr’ can be used as general terms also.
Problem based on General Terms
Type : 1_ Q4 Find the 7th term in the expansion of
[4x – (1 / 2vx)]13
Sol : T7 = T6+1 = 13C6(4x)13-6 – (1/2vx) 6
= 13C6.47x7. 1 /(26.x3)
= 13C6. 28.x4
= 13!/ (6!x7!) . 28. x4
= T7 = 439296x4
Type II : Find the coefficient of x-7 in the expansion of (ax – 1/ bx2) 11
Sol.: General Term , Tr+1 = 11Cr(ax)11-r – (1/ bx2 ) r
Tr+1 = (-1)r 11Cr. (a11-r / br) x11-3r ————–> (i)
Putting 11 – 3r = -7
Or 3r = 18
r = 6
From (i) to T7 = (-1)6 11C6. ( a5 / b6) x-7 ————–> (i)
Hence, the coefficient of x-7 in ax- (1 / b x2) 11 is 11C6a5b-6
Type III : Find the term independent of ‘x’ in [(3 x2 / 2) – (1/ 3x) ] 9
Sol.: General Term, Tr+1 = 9Cr (3 x2 / 2) 9-r – (1/3x) r
= (-1)r 9Cr ( 3/2) 9-r x18-2r (1 / 3r. xr )
Tr+1 = (-1)r9 Cr (39-2r / 29-r). x18-3r ——-> (i)
Putting 18- 3 r = o
r = 6
So, from (i), 7th term is independent of ‘x’, and its value is:
T7 = (-1)6 . 9C6. (3-3 / 23) xo
= 9 ! /(6! X 3!) . 1/ (33 x 23)
= T7 = (7/18)
Pth term from end:
‘P’th term from end in the expansion of (x+y)n is (n-P+2)th term from beginning.
Ex.: Find the 4th term from the end in the expansion of [ (x3/2) – (2/x2) ] 7
Sol.: 4th term from end = (7-4+2)th or 5th term from beginning.
T5 = T4+1 = 7C4 (x3/2)7-4 . (-2/x4) 4
= 7C4 (x3 /2) 3 ( -2/ x2) 4
= 7! / (4! X 3!) . (x9/8) . (16/ x8)
= (7.6.5 / 3.2.1) .2x
T5 = 70x
Hence ‘4’ term, from the end = 70x.
Middle Terms: It depends upon the value of ‘n’.
Case -1 : When ‘n’ is even, then total number of terms in (x+y)n is odd. So there is only one middle term i.e. [(n/2) + 1] th them is the middle term.
So we find (Tn+1/2). th term in this case, if ‘n’ is even.
Case II : When ‘n’ is odd, then total number of terms in (x+y)n is even. So there are two middle terms i.e. (n+1) /2 th and (n+3) /2 th are true middle terms.
so we find T(n+1)/2 th and T(n+3)/2 th in this case if ‘n’ is odd.
Ex.: Find the middle – term in the expansion of [ 3x – (x3 / 6)]9
Sol.: Here total no. of terms are 10 (even). So there are true middle-terms
i.e (9+1) / 2 th and (9+3) / 2 th. So we have to find – out ‘T5’ and ‘T6’.
T5 = T4+1 = 9C4(3x)9-4 (-x3 / 6) 4
= 9! / (4! X 5!) .35 x5 ( x12 / 64)
= (9.8.7.6 / 4.3.2.1) 35 / (24 x 34) x17
T5 = (189 / 8) x17
T6 = T5+1 = 9C5(3x)9-5 (-x3 / 6) 5
= 9! / (5! X 4!) .34 x4 (x15 / 65)
= -(9.8.7.6 / 4.3.2.1) 34(25 x 35) x19
T6 = – ( 21 / 16) x19
Greatest – term in (1+x)n : If ‘Tr’ and ‘Tr+1’ be the ‘r’ th and (r+1)th terms in the
Expansion of (1+x)n, then :
Tr+1 = nCr(1)n-r xr = nCr xr
And Tr = nCr-1. xr-1
So: Tr+1 / Tr = (nCr xr / nCr-1 xr-1) = (n-r+1)/r |x|
If ‘Tr+1 be the greatest term, then Tr+1 Tr
Or Tr+1 / Tr 1
since (n-r+1) / r. |x| >=1, where ‘r’ is a ‘+’ ve integer.
This inequality, changes either to the form r<=m+f pr r <= m, where ‘m’ is a ‘+’ ve integer and ‘f’ is a fraction. So we get:
r <= m + f —————> (i)
or r <= m ——————> (ii)
In case (i), ‘T’m+1 is the greatest term, and in case (i) ‘T’m and ‘T’m+1 are the greatest terms, and both are equal.
Short-cut: First calculate m = | x (n+1) / (x + 1) |
Case (1) If ‘m’ is an integer, then ‘T’m and ‘T’m+1 are the greatest terms and both are equal.
Case (2) If ‘m’ is not an integer, then T[m]+1 will be the greatest term, where [.] denotes greatest integer function.
Ex.: Find numerically the greatest term in the expansion of (2+3x),
when x = (3 / 2)
Sol.: 1 Method : (2+3x)9 = 29 [1+ 3x / 2] 9
In the expansion of [(1 + 3x) / 2] 9, we have :
Tr+1 / Tr = (9-r+1)/ r |3x / 2|
= ((10 – r)/r) | (3/2) x(3/2) | 3
= (10 – r) / r x 9 / 4
Tr+1 / Tr = (90- 9r) / 4r
Putting Tr+1 / Tr >= 1
(90-9r) / 4r >= 1
or 90 >= 13 r
or r 90 / 13
or r <= 6 + 12 / 13
T6+1 or ‘T7’
is the greatest term.
‘T7’ in [1 + (3x / 2)] 9
T7 = T6+1 = 9C6 (3x / 6) 6
= 9! / (3! X 6!) .[ (3 / 2) x (3 / 2)] 6
= (9 .8.7 / 3.2.1) x (96 / 46)
= (3 x 7 x 96) / 45 = (3 x 7 x 312) / 210
= 7. (313 / 210)
So greatest term in (2+ 3x)9 is :
= 29. 7. (313 / 210)
= (7 x 313) / 2
II- Method : (2+3x)9 = 29 [(1 + 3x) / 2] 9
= 29 [1 + 9 / 4] 9
since x = 3 / 2
Here m = | x (n + 1) / (x + 1)| = | 9/4 (9+1) / 9/4 + 1|
= 90 / 13
So greatest term in the expansion is T[m]+1 = T3+1 = T7
Now the method is same as in method (1)
Greatest Coefficient : In any binomial expansion middle-term has the greatest.
Coefficient. So
(i) If ‘n’ is even, then greatest – coefficient = nCn/2
(ii) If ‘n’ is odd, then greatest – coefficients are nC(n+1)/ 2 and nC (n-1)/2
Properties of Binomial coefficients :
(1) The sum of binomial coefficient in (1 + x)n is 2n.
Proof (1 + x)n = Co+C1x+C2x2 + —– + Cnxn———–> (i)
Putting x = 1 :
2n = Co + C1 + C2 + ———– + Cn ———–> (ii)
Ex.: Prove that the sum of the coefficients in the expression (1+x – 3x2)2163
is ‘-1’.
Sol.: Putting x = 1 in (1 + x – 3x2)2163
Some of the coefficients = (1 + 1 – 3)2163
= (-1)2163 = -1
(2) The sum of the coefficients of the odd-terms in (1+x)n is equal to the sum of coefficients of the even terms and each is equal to 2n-1.
Proof: Putting x = -1, in eg(1) :
O = Co – C1 + C2 – C3 + —— + (-1)nCn
and from (ii): 2n = Co + C1 + C2 + ——— + Cn
Adding these egn:
2n = 2 ( Co + C2 + C4 + —————)
or Co + C2 + C4 + ——- = 2n-1 ————> (ii)
Subtracting these egn:
2n = 2 (C1 + C3 + C5 + ————–)
or C1 + C3 + C5 + ——- = 2n-1 ————> (iv)
From (iii) and (iv) :
C0 + C2 + C4 + ——- = C1 + C3 + C5 + ——- = 2n-1
Ex.: Evaluate the sum of the 8C1 + 8C3 + 8C5 + 8C7
Sol.: since nC1 + nC3 + nC5 + nC7 + ——– = 2n-1
Here n = 8
8C1 + 8C3
+ 8C5 + 8C7 =
(28-1)
= 27
= 128
( 8C9, 8C11 etc. are not possible)
Some important results:
(i) In the expansion of (1+x)n, coefficient of xr = nCr
(ii) In the expansion of (1-x)n, coefficient of xr = (-1)r. nCr
(iii) If ‘n’ is a negative integer or fraction, then
(1+x)n = 1 + (n / 1!) x + [ n (n-1)/ 2!] x2 + [n(n-1)(n-2) / 3!] x3 + ————-
+ [n(n-1)(n-2) ———–(n-r+1) / r!]xr
+ ————–
Here | x | <1, i.e. – 1<x<1 is necessary for its validity.
(iv) In (1+x)n, general – term Tr+1 = [n(n-1)(n-2) ————-(n-r+1) / r!]. x2
(v) nCr + nCr-1 = n+1Cr
(vi)
nCx = nCy x = y or x +
y = n
(vii) nCr = n/ r. n-1Cr-1
Multinomial theorem : (For a ‘+’ve integral index):
If nN, and x1, x2, x3, ——–xm C, then
(x1 + x2 + x3 + ———+xm)n = ? n! / (n1! n2! —nm!) x1n1, x2n2 ….xmnm
Where n1, n2, n3 ——–, nm are non-negative integers, satisfying the condition
n1 + n2 + ———–+nm = n
Note: The coefficient of x1n1. x2n2. ———xmnm in the expansion of
(x1 + x2 + x3 + ——————- + xm)n is :
= n! / (n1! x n2! —nm!)
So, general-term in (a+b+c+d)n = n! / (p! x q! x r! x s!). ap.bq.cr.ds.
Where p+q+r+s = n, and p, q, r, s W.
(2) Number of terms in (x1 + x2 + x3 + ——— + xm)n : n+m-1Cm-1.
Ex.: Find the number of terms in the expansion of (2x – 3y + 4z)100
Sol.: Number of terms = 100+3-1C3-1 = 102C2
= 102 ! / (2! X 100!)
= (102 x 101) / (2 x 1) = 5151
General term of a multinomial – theorem :
Tr+1 = n! / (n1! x n2! —nm!) x1n1. x2n2 ———–xmnm
EXAMPLES
Q1. Find the coefficient of x3 y4 z2 in the expansion of (2x – 3y + 4z)9.
Sol. General Term in (2x – 3y + 4z)9
= 9! / (n1! X n2! X n3!). (2x)n1. (-3y)n2. (4z)n3
= 9! / (n1! X n2! X n3!). 2n1 (-3)n2. (4)n3. xn1. yn2. zn3
Putting n1 = 3, n2 = 4, n3 = 2 :
= 9! / (3! x 4! x 2!). 23 (-3)4. (4)2. x3 y4 z2
= [ 9. 8. 7. 6. 5. 4!/ (3.2.1. 4!.2)] x 8 x 81 x 16 x3 y4 z2
Coefficient of x3 y4
z2 = 9 x 8 x 7 x 5 x 8 x 81 x 8
= 13063600
Greatest coefficient in the expansion of (x1 + x2 + ——– + xm)n is
= n! / (q!) m-r ( q+1!) r
Where ‘q’ is the quotient and ‘r’ is the remainder, when ‘n’ is divided by ‘m’.
Ex.: Find the greatest coefficient in the expansion of (a + b + c + d) 15.
Sol.: Here n = 15, m = 4
15/4 is quotient 3 and remainder 3.
since q = 3 and r = 3
Hence greatest – coefficient = 15! / [(3!) 4-3 x (3+1!)3]
= 15! / [(3!) x (4!)3 ]
= 15! / (3! x 4! x 4! x 4!)
Ex.: Find the coefficient of x7 in the expansion of (1+3x-2x3)10.
Sol.: General term in (1+3x-2x3)10
= 10! / (n1! x n2! x n3!). (1)n1 (3x)n2 (-2x3)n3
= 10! / (n1! x n2! x n3!). 3n2 (-2)n3 xn2+3n3
Where n1 + n2 + n3 = 10 ————–> (i)
For coefficient of x7 : n2 + 3n3 = 7 ————-> (ii)
From (ii), possible non-negative integral values of ‘n2’ and ‘n3’ are :
n2 = 7, n3 = o since from (i) : n1 = 3
n2 = 1, n3 = 2 since from(i) : n1 = 7
or n2 = 4, n3 = 1 since from (i): n1 = 5
So required coefficient of x7 :
10! / (3! x 7! x 0!) . (3)7 (-2)0 + 10! / (7! x 1! x 2!). (3)1(-2)2 + 10! /(5! x 4! x 1!). 34 (-2)1
(10. 9. 8 7!) / (7!.3.2.1).37 + (10. 9. 8. 7!) / (7! . 2) x 3 x 4 – [(10. 9. 8. 7. 6. 5! ) / (5!. 4.3.2.1.)] .3. 2.
= 10 x 9 x 4 x 36 + 10 x 9 x 4 x 3 x 4 – 10 x 9 x 7 x 6 x 33 x 2
= 10 x 9 x 4 (36 + 12 – 7 x 34)
= 360 x (729 + 12 – 567)
= 62640
Some tips on the solution of binomial – coefficients:
(1) If the difference of the lower suffixes of binomial coefficients in each term is same.
For Ex.: C1 C3 + C2 C4 + C3 C5 + —— etc.
Then :
Case -1 : If each term is positive, then
(1+x)n = C0 + C1x + C2x2 + ———— Cn xn —————–> (i)
Interchanging ‘1’ and ‘x’:
(x+1)n = C0 xn + C1 xn-1 + C2 xn-2 + ——— + Cn ————->(ii)
Then multiplying (i) and (ii), and equate the coefficient to suitable power of ‘x’ on both sides.
Case –II : If terms are alternately positive and negative
Then:
(1-x)n = C0 – C1 x + C2 x2 – ————– + (-1)n Cn xn —————> (1)
and (x+1)n = C0 xn + C1 xn-1 + C2 xn-2 + ———- + Cn ————-> (2)
The multiplying (1) and (2), and equate the coefficient of suitable power of ‘x’ on both sides.
Note : [ (Odd – number) / 2] = 8
(2) If the sum of the lower suffixes of binomial – coefficients in each term is same.
For Ex.: C0 Cn + C1 Cn-1 + C2. Cn-2 + ——- + Cn C0
Then:
Case – 1 : If each term is positive, then
(1+x)n = C0 + C1 x + C2 x2 – ————– + Cn xn —————> (1)
and (1+x)n = C0 xn + C1 x + C2 x2 + ———- + Cn xn————-> (2)
Then multiplying (i) and (ii), and equate the coefficient of suitable power of ‘x’ on both sides.
Case –II : If terms are alternately positive and negative,
The (1+x)n = C0 + C1 x + C2 x2 – ————– + Cn xn —————> (1)
and (1-x)n = C0 – C1 x + C2 x2 + ———- + (-1)n Cn xn————-> (2)
Then multiplying (i) and (ii) and equating the coefficient of suitable power of ‘x’ on both side.
PROBLEMS
(1) Show that the middle – term in the expansion of (1+x)2n is
1. 3. 5 ——- (2n-1) / (n!) . 2n xn, ‘n’ being a positive integer.
Sol.: The no. of terms in (1+x)2n = 2n+1 (odd).
It’s ,middle-term = (2n + 1) / 2 = (n+1)th term.
Tn+1
= 2nCn xn
= 2n! / (n! x n!). xn
= 2n (2n-1) —— 4.3.2.1 / (n! x n!). xn
= [{(2n-1) (2n-3) —– 3.1.} { 2n (2n-2) —— 4.2.}] / (n! x n!). xn
= [{1. 3. 5. —- (2n-1)} 2n {1.2 —- n}] / (n! x n!) . xn
= [{1.3.5—-(2n-1)}. 2n] / (n! x n!). xn
= Tn+1 = 1. 3. 5 – (2n-1) / (n!). 2n xn
(2) Find the term independent of ‘x’ in the expansion of
(i) (1+x+2x3) [(3 x2 / 2) – (1/3x)] 9
(ii) [( x1/3 / 2) + x-1/5] 8
Sol.: (i) (1+x+2x3) [(3/2)x2 – (1/3x)] 9
= (1+x+2x3) { [(3/2)x2] 9 – 9C1 [(3/2)x2 ] 8 1/3x + ———- +
+ 9C6 [(3/2)x2] 3 (1/3x)6 – 9C7 [(3/2)x2] 2 (1/3x)7 —}
= (1+x+2x3) { [(3/2)x2 ] 9 – 9C1 (37 / 28)x15 + —- + 9C6 (1 x 1 / 23 x 33) – 9C7 1/ (22 x 35) 1/ x3 + —-}
Term independent of ‘x’ :
9C6 x 1/ (23 x 33) – 9C7 2 / (22 x 35)
= 9! / (6! x 3!) . 1/ (8 x 27) – 9!/(7! x 2!) . 1/ (2 x 243)
= (9. 8. 7. 6!) / ( 6!. 3. 2. 1.) x 1/(8. 27) – (9. 8. 7!) / (7! . 2). 1/ (2.243)
= 7 / 18 – 2 / 27 = 17 / 54
(ii) [(1 / 2) x1/3 + x-1/5] 8
Sol.: General Term Tr+1 = nCr [(1/2) x1/3] n-r. (x-1/5) r
n-r -r
= nCr [(1/2) n-r] x 3 x 5
Here n = 8
= 8Cr (1/2) 8-r x (8-r)/3 -r/5
40 -8r
Tr+1 = 8Cr (1/2) 8-r x 15 —————> (i)
Putting (40 – 8r) / 15 = 0, we have r = 5
From (i), Term independent of ‘x’
:
T6 = 8C5 (1/2) 8-5
= 8! / (5! X 3!) . 1 / 23
= (8. 7. 6. !5) / (5!. 3.2.1) . 1 / 8
= T6 = 7
(3) Find the coefficient of ‘x’ in the expansion of (1-2x3 + x5) [1 + (1/x)]8
Sol.: (1-2x3 + 3x5) [1 + (1/x)] 8
= (1-2x3 + 3x5) [1 + 8C1 (1/x) + 8C2 (1/ x2) + 8C3 (1/ x3 ) + 8C4 (1/ x4)+ 8C5 (1/ x5 )+ — + 8C8 (1/ x8)
coefficient of x = -2. 8C2 + 3 8C4
= -2. 8! / (2! x 6!) + 3. 8! / (4! x 4!)
= -2. (8. 7) / 2 + 3 (8. 7. 6. 5.) / (4.3.2.1)
= -56 + 210
= 154
(4) Prove that the ration of the coefficient of x10 in (1-x2)10 and the term independent of ‘x’ in [x – (2/x)] 10 is 1 : 32.
Sol.: In (1-x)2 : Tr+1 = 10Cr (-1)r (x2)r
Putting r = 5
T6 = –10C5 x10
Coefficient of x10 = –10C5
In [x – (2/x)] : Tr+1 = 10Cr (-1)r (x)10-r (2/x)r
= (-1)r 10Cr. 2r. x10-2r
Putting 10 – 2r = 0
r = 5
So term independent of x : T6 = (-1)5 10C5. 25
Hence their ratio = (-10C5) : (-32. 10C5)
= 1 : 32
(6) If third term in the expansion of (x + x logx)5 is 10,00,000. Find the value of ‘x’.
Sol.: Putting log10x = z in the given expression :
We have :
( x + xz)5
T3 = T2+1 = 5C2 (x)5-2 (xz)2
= 5C2 x3. x2z
= 5! / (2! x 3!) x2z+3
= (5 x 4) / 2! x2z+3
= T3 = 10x2z+3
10,00,000 =
10. x2z+3
Or x2z+3 = 105
(10z)2z+3
= 105
or 102z2+3z = 105
2z2 +
3z = 5
[Log10x = z]
or 2z2 + 3z – 5 = 0
or (z-1) (2z+5) = 0
z
= 1, – 5 / 2
or log10x = 1 or log10x = – 5 / 2
since x = 10 or 10-5/2
(7) If in the expansion of (1+x)m (1-x)n, the coefficients of ‘x’ and ‘x2’ are ‘3’ and ‘-6’ res. Find the value of ‘m’.
Sol.: (1+x)m (1-x)n = [mC0 + mC1x + mC2x2 + —- + mCm xm]
[nCo – nC1x + nC2x2 + ——- + (-1)n nCnxn]
Coefficient of x = mC1 x nCo – mC0. nC1
= m! / ( 1! x m-1!) x 1 – 1 x n! / (1! x n-1!)
= m – n = 3 —————> (i)
Coefficient of x2 = –mC1 x nC1 +nC0 x mC2 + mC0 x nC2
= – m!/ (1! x m-1!) x n! / (1! x n-1!)+ 1 x m! / (2! x m-2!) + 1 x n! / (2! x n-2!)
= -mn + m (m-1) / 2 + n(n-1) / 2 = -6
or – 2mn + m(m-1) + n(n-1) = -12
or -2mn + m2 – m + n2 – n = 12
or (m-n)2 – (m+n) = -12
From (i), putting the value of (m-n) :
– 9 + (m + n) = 12
or m + n = 21 ———–> (ii)
egn (i) + egn(ii) = 2m = 24
m = 12
Q8. If the coefficients of (2r + 1)th term and (r + 2)th term in the expansion of (1 + x)43 are equal, find ‘r’.
Sol.: In (1 + x)43 : T2r+1 = 43C2r. x2r
Coefficient = 43C2r
And Tr+2 = 43Cr+1 xr+1
Coefficient = 43Cr
According to the questions:
43C2r = 43Cr+1
2r + r + 1 = 43
or 3r = 42
r = 14
Q9. If the coefficient of ‘4’th and ‘13’th terms in the expansion of [x2 + (1/x)] n be equal, then find the term which independent of ‘x’.
Sol.: T4 = T3+1 = nC3 (x2)n-3. 1/ x3
Coefficient = nC3
T13 = T12+1 = nC12 (x2)n-12 1 / x12
Coefficient = nC12
According to the question:
nC3 = nC12
n =
12 + 3
n = 15
Expansion = [x2
+ (1/x)]15
Now Tr+1 = 15Cr. (x2)15-r. 1/ xr
Tr+1 = 15Cr. x30-3r ————-> (i)
Putting :
30 – 3r = 0
r = 10
From (i) T11 = 15C10 = 15!/(10! x 5!) = (15 x 14 x 13 x 12 x 11) / (5 x 4 x 3 x 2 x 1)
= 3003.
Q10. In the expansion of (a – b)n, n <= 5, if the sum of the 5th and 6th terms is zero. Find ( a / b) in terms of ‘n’.
Sol.: T5 = T4+1 = nC4 (a)n-4 (-b)4
T5 = nC4 an-4 b4
T6 = T5+1 = nC5 (a)n-5 (-b)5 = –nC5 an-5 b5
T5 + T6 = 0
nC4
an-4 b4 – nC5
an-5 b5 = 0
or nC4 an-4 b4 = nC5 an-5 b5
or n!/(4! x n-4!) an-4 = n!/(5! x n-5!) an-5 b
or an-4 / (n-4)(n-5!) = an-5 / 5(n-5!) b
or an-4 / an-5 = b / 5 (n-4)
or a(n-4)-(n-5) = (n – 4) / 5 .b
or a = (n – 4)/5 . b
or a/b = (n – 4) / 5
Q11. Find the coefficient of xr in the expansion of [x + (1/x)] n, if it occurs.
Sol.: General term : Tp+1 = nCp (x)n-p (1/x) p
Tp+1 = nCp xn-2p —————> (i)
Putting n-2p = r
p =
(n – r) / 2
From: (i) T (n-r) / 2 +1 = nC(n-r) / 2 xr
Coefficient of xr = nC (n – r) / 2
Q12.: Prove that the coefficient of the term independent of ‘y’ in the expansion of
[(y + 1)/( y 2/3 – y1/3 + 1) – (y – 1) / (y – y1/2)]10 is 210.
Sol.: We have (y + 1) / (y 2/3 – y1/3 + 1)
Putting y = t3, we have
= (t3 + 13) / (t2 – t + 1) = (t + 1) (t2 – t + 1) / (t2 – t + 1)
= t + 1
(y + 1) / (y2/3 – y1/3 + 1) = y1/3 + 1
and Putting y = a2 in (y – 1) / (y – y1/2 ) :
= (a2 – 1) / (a2 – a) = (a+1) (a-1) / [a (a-1)]
= (a + 1) / a = 1 + 1 / a
(y – 1) / (y – y1/2)
= 1 + 1 /vy
(y + 1) / (y2/3
– y1/3 +1) – (y –
1) / (y – y1/2)]10
= [y1/3 + 1 – 1
– (1/ y1/2)] 10
= (y1/3 – y-1/2 )10
In ( y1/3 – y-1/2)10,
Tr+1 = 10Cr (y1/3)10-r. (-y-1/2)r
= (-1)r 10Cr. (10-r) / 3 – r/2
Tr+1 = (-1)r 10Cr. y(20-5r) / 6
Putting (20 – 5r) / 6 = 0
or r = 4
Putting this value in (1) T5 = (-1)4 10C4
= 10!/ (6! x 4!) = (10 x 9 x 8 x 7) / (4 x 3 x 2 x 1)
T5 = 210
Q13: x4-r occurs in the expansion of [x + (1/ x2)] 4n, prove that its coefficients is:
= (4n!) /[ (4/3)n-r]! x [(4/3)(2n+r)]!
Sol.: In [x + (1/x2)]4n, Tp+1 = 4nCp (x)4n-p (1/ x2)p
Tp+1 = 4nCp x4n-3p ———-> (i)
Putting :
4n – 3p = 4r
or 4 ( n-r ) / 3 = p
From (i) Tp+1 = 4nC4(n-r) / 4. x4r
Coefficient of x4r = 4nC4 (n-r) / 3
= (4n!) / [(4/3)n-r]! x [(4n/1) – 4(n-r)/3]!
= (4n!) / [(4/3)n-r]! x [(4/3) 2(n+r)]!
Q14. Find the coefficient of x50 in (1+x)41 (1-x+x2)40.
Sol.: (1+x)41 (1-x+x2)40 = (1+x) (1+x)40 (1-x+x2)40
= (1+x) [ (1+x) ( 1-x + x2)]40
= (1+x) (1+x3)40
General Term = Tr+1 = (1+x) [40Cr (x3)r]
= 40Cr (1+x) x3r
= 40Cr (x3r + x3r + 1)
Here either 3r = 50 or 3r+1 = 50
r = (50 / 3) or (49 / 3)
The value of ‘r’ is a fraction, so it doesn’t contain the term x50. So coefficient of x50 is ‘0’.
Q15.: Show that that the term independent of ‘x’ in the expansion of
[x + (1/x)] 2n is [1. 3. 5. —- (2n-1) / (n!)] 2n
Sol.: General Term Tr+1 = 2nCr (x)2n-r (1/x) r
= 2nCr. x2n-2r ———> (i)
Here 2n – 2r = 0
or n = r
From (i) Tr+1 = 2nCn
= 2n! / (n! x n!)
= [2n (2n-1 ) —— 3. 2. 1] / ( n! x n!)
= { 2n (2n-2) —- 4. 2 } { (2n-1) (2n-3) —– 3.1.} / (n! x n!)
= [2n {n (n-1) —–2.1.}] { (2n-1) ——- 4.3.1.} / (n! x n!)
= 2n. n!{(2n-1) —- 5. 3. 1. / (n! x n!)
= {1. 3. 5. —– (2n -1)} 2n / n!
Q16. The 3rd, 4th and 5th terms in the expansion of (x+a)n are respectively ‘84’, ‘280’ and ‘560’, find the value of ‘x’, ‘a’ and ‘n’.
Sol.: Tr+1 = nCr xn-r. ar
Putting r = 2, 3 and 4 respectively
T3 = nC2 xn-2. a2 = 84 ————>(i)
T4 = nC3 xn-3 a3 = 280———->(ii)
and T5 = nC4 xn-4 a4 = 560 ——–>(iii)
egn (i) x egn(iii) : [nC2 xn-2 a2] [nC4 xn-4 a4] = 84 x 560
= n!/[2! x (n-2)!] x n! / [4! x (n-4)!] . x2n-6 a6 = 84 x 560
or n (n-1) / 2 x n(n-1) (n-2) (n-3) / 4! x x2n-6 a6 = 84 x 560 ——-> (iv)
Squaring of egn (ii), we have :
(nC3 x n-3 a3)2 = 2802
nC3
x nC3 x x2n-6
x a6
= 2802
= n! / [3! x (n-3)!] x n! / [3! x (n-3)!] x x2n-6 a6 = 2802
or n (n-1)(n-2) / 6 x n(n-1) (n-2) (n-3) / 3! x x2n-6 a6 = 280 x 280 ——-> (v)
egn (v) egn(iv) :
n2 (n-1)2 (n-2)2 /
(6 x 3!) x 2 x 4! / [n2(n-1)2
(n-2)(n-3)] = (280 x 280) / (84 x 560)
or 4 (n-2) / 3 (n-3) = 5 / 3
or 4n – 8 = 5n – 15
n = 7
Putting this value in (i), (ii) and (iii) :
7C2 x5 a2 = 84 —————> (vi)
7C3 x4 a3 = 280 —————->(vii)
7C4 x3 a4 = 560—————–>(viii)
egn (vii) egn(vi):
(7C3 x4 a3) / (7C2 x5 a2) = 280 / 84
[7! / (3! x 4!)a] / [7! / (2! x 5!)x] = 10 / 3
or 7! / (3! x 4!) x (2! x 5!) / 7! x a / x = 10 / 3
or 5 / 3 x a / x = 10 / 3
or a = 2x
Putting this value in egn (vi):
7C2. x5. 4x2 = 84
or 7! / (2! x 5!)x7 = 21
(7
x 63) / 2 x7
=
21
x7 = 1
x = 1
Putting this value in (ix) = a = 2
Q17. Let ‘n’ be a positive integer. If the coefficients of second, third and fourth terms in (1+x)2 are in arithmetic progression, then find the value of ‘n’.
Sol: General Term : Tr+1 = nCr xr
2nd
Term : T2 = nC1 x
Coefficient = nC1
3rd
Term : T3 = nC2 x2
Coefficient = nC2
Similarly coefficient of 4th term = nC3
These are in A. P., so.
2 nC2 = nC1 + nC3
2 [n! / {2! x
(n-2)!}]
= n! / {1! x
(n-1!)} + n! / {3! x (n-3!)}
or n! / (n-2!) = n! [1 / (n-1!) + 1 / {6 (n-3!}]
or 1/ [(n-3) x (n-3!) = 1 / [(n-1)x (n-2) x(n – 3!)] +1/ [6! (n-3!)] )
or 1 / (n – 2) – 1 / [(n-1) (n-2)] = 1 / 6
or (n – 1 – 1) / [(n-1) (n-2)] = 1 / 6
or (n- 2) / [(n-1) (n-2)] = 1 / 6
or n – 1 = 6
n = 7
Q18. The 6th term in the expansion of [(1/ x8/3) + x2 log10x]8 is 5600. Prove that x =10.
Sol.: T6 = T5+1 = 8C5 (1/ x8/3) 8-5 ( x2 log10x ) 5
or 8C5 x (1 / x8) x c10 x (log10x)5 = 5600
8! / (5! x 3!)
x c2 (log10x)5
= 5600
8. 7. 6. / 6 x
c2 (log10x)5
= 5600
or x2(log10x)5 = 100 = 102
Clearly x = 10 satisfied as log1010 = 1.
If x > 10 or < 10, the result will change in inequality.
———–x——————–x————————-x———————–x—————-x—–