Binomial Theorem Tutorial

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 + ⁿC₁x^n-1y+ⁿC₂x^n-2y²+-----+ⁿC_rx^n-ry^r+ -------+---------+ⁿC_n-1xy^n-1+ⁿc_nyⁿ.

It can be represented as:

(x+y)ⁿ = ⁿC_rx^n-ry^r

Particular – Cases :

(i) Replacing ‘y’ by ‘-y’, we have :

(x-y)ⁿ = ⁿC_oxy^o-nC₁x^n-1y+ⁿC₂x^n-2y²-------+(-1)^{r
n}C_rx^n-ry^r+------+(-1)ⁿ ⁿC_nx^oyⁿ.

It can be represented as :

(x+y)ⁿ = (-1)^r ⁿC_rx^n-ry^r

(ii) Replacing ‘x’ by ‘1’ and ‘y’ by ‘x’, we have :

(1+x)ⁿ = ⁿC_ox^o+ⁿC₁x+ⁿC₂x²+---------+ⁿC_rx^r+------+ⁿC_n-1x^n-1+ⁿC_nxⁿ.

or = ⁿC_rx^r

(ii) Replacing ‘x’ by ‘-x’, we have :

(1+x)ⁿ = ⁿC_ox^o-ⁿC₁x¹+ⁿC₂x² - ---------+(-1)^{r n}C_rx^r+------+ⁿC_n-1(-1)^n-1+(-1)ⁿ ⁿC_nxⁿ.

or = (-1)^rnC_rx^r

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 ⁿC_r = ⁿC_n-r,

so we have :

ⁿC_o = ⁿC_n

ⁿC₁=ⁿC_n-1

ⁿC₂=ⁿC_n-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+ⁿC₁x^n-1a + ⁿC₂x^n-2a²+ⁿC₃x^n-3a³ + ------- + ⁿC_naⁿ

= (xⁿ+ⁿC₂x^n-2a² + ---------) + (ⁿC₁x^n-1a+ⁿC₃x^n-3a³+ ------)

(x+a)ⁿ = P+Q ------------------------> (1)

and (x-a)ⁿ = xⁿ-ⁿC₁x^n-1a+ⁿC₂x^n-2a²-ⁿC₃x^n-3a³ + ----+ (-1)^{n n}C_naⁿ

= (xⁿ+ⁿC₂x^n-2a² + ----) – (ⁿC₁x^n-1a+ⁿC₃x^n-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

T_r+1 = ⁿC_rx^n-ry_r

Explanation: We know

(x-y)ⁿ = ⁿC_oxⁿy^o+ⁿC₁x^n-1y¹+ⁿC₂x^n-2y²+----+ⁿC_nx^oy^o

Here:

First term T₁ = ⁿC_oxⁿy^o

T₂ = ⁿC₁x^n-1y¹

T₃ = ⁿC₂x^n-2y²

------------------------

T_r = ⁿC_r-1 x^n-(r-1)y^r-1

Putting r = r+1 in this expression, we get:

General Term: T_r+1 = ⁿC_rx^n-ry^r

Note : ‘T_r’ can be used as general terms also.

Problem based on General Terms

Type : 1_ Q4 Find the 7^th 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 , T_r+1 = ¹¹C_r(ax)^11-r - (1/ bx² ) ^r

T_r+1 = (-1)^r ¹¹C_r. (a^11-r / b^r) x^11-3r --------------> (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, T_r+1 = ⁹C_r (3 x² / 2) ^9-r – (1/3x) ^r

= (-1)^r ⁹C_r ( 3/2) ^9-rx^18-2r (1 / 3^r. x^r )

T_r+1 = (-1)^r9C_r(3^9-2r / 2^9-r). x^18-3r -------> (i)

Putting 18- 3 r = o

r = 6

So, from (i), 7^th term is independent of ‘x’, and its value is:

T₇ = (-1)⁶ . ⁹C₆. (3^-3 / 2³) x^o

= 9 ! /(6! X 3!) . 1/ (3³ x 2³)

= T₇ = (7/18)

P^th 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 4^th term from the end in the expansion of [ (x³/2) - (2/x²) ] ⁷

Sol.: 4^th term from end = (7-4+2)th or 5^th 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 (T_n+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!) .3⁵x⁵(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!) .3⁴x⁴ (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 ‘T_r’ and ‘T_r+1’ be the ‘r’ th and (r+1)th terms in the

Expansion of (1+x)ⁿ, then :

T_r+1 = ⁿC_r(1)^n-r x^r = ⁿC_rx^r

And T_r = ⁿC_r-1. x^r-1

So: T_r+1/ T_r = (ⁿC_r x^r / ⁿC_r-1 x^r-1) = (n-r+1)/r |x|

If ‘T_r+1 be the greatest term, then T_r+1 ³ T_r

Or T_r+1 / T_r ³ 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 :

T_r+1/ T_r= (9-r+1)/ r |3x / 2|

= ((10 – r)/r) | (3/2) x(3/2) | ³

= (10 – r) / r x 9 / 4

T_r+1/ T_r = (90- 9r) / 4r

Putting T_r+1/ T_r>= 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 = ⁿC_n/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 = C_o+C₁x+C₂x² + ----- + C_nxⁿ-----------> (i)

Putting x = 1 :

2ⁿ = C_o + C₁ + C₂ + ----------- + C_n -----------> (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 2^n-1.

Proof: Putting x = -1, in eg(1) :

O = C_o – C₁ + C₂ – C₃ + ------ + (-1)ⁿC_n

and from (ii): 2ⁿ = C_o + C₁ + C₂ + --------- + C_n

Adding these egⁿ:

2ⁿ = 2 ( C_o + C₂ + C₄ + ---------------)

or C_o + C₂ + C₄ + ------- = 2^n-1 ------------> (ii)

Subtracting these egⁿ:

2ⁿ = 2 (C1 + C3 + C5 + --------------)

or C₁ + C₃ + C₅ + ------- = 2^n-1 ------------> (iv)

From (iii) and (iv) :

C₀ + C₂ + C₄ + ------- = C₁ + C₃ + C₅ + ------- = 2^n-1

Ex.: Evaluate the sum of the ⁸C₁ + ⁸C₃ + ⁸C₅ + ⁸C₇

Sol.: since ⁿC₁ + ⁿC₃ + ⁿC₅ + ⁿC₇ + -------- = 2^n-1

Here n = 8

⁸C₁ + ⁸C₃ + ⁸C₅ + ⁸C₇ = (2^8-1)

= 2⁷

= 128

( ⁸C₉, ⁸C₁₁ etc. are not possible)

Some important results:

(i) In the expansion of (1+x)ⁿ, coefficient of x^r = ⁿC_r

(ii) In the expansion of (1-x)ⁿ, coefficient of x^r = (-1)^r. ⁿC_r

(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!]x^r + --------------

Here | x | <1, i.e. – 1<x<1 is necessary for its validity.

(iv) In (1+x)ⁿ, general – term T_r+1 = [n(n-1)(n-2) -------------(n-r+1) / r!]. x2

(v) ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r

(vi) ⁿC_x = ⁿC_y x = y or x + y = n

(vii) ⁿC_r = n/ r. ^n-1C_r-1

Multinomial theorem : (For a ‘+’ve integral index):

If nÎN, and x₁, x₂, x₃, --------x_m Î C, then

(x₁ + x₂ + x₃ + ---------+x_m)ⁿ = ? n! / (n₁! n₂! ---n_m!) x₁ⁿ¹, x₂ⁿ² ….x_m^nm

Where n₁, n₂, n₃ --------, n_m are non-negative integers, satisfying the condition

n₁ + n₂ + -----------+n_m = n

Note: The coefficient of x₁ⁿ¹. x₂ⁿ². ---------x_m^nm in the expansion of

(x₁ + x₂ + x₃ + ------------------- + x_m)ⁿ is :

= n! / (n₁! x n₂! ---n_m!)

So, general-term in (a+b+c+d)ⁿ = n! / (p! x q! x r! x s!). a^p.b^q.c^r.d^s.

Where p+q+r+s = n, and p, q, r, s Î W.

(2) Number of terms in (x₁ + x₂ + x₃ + --------- + ^xm)ⁿ : ^n+m-1C_m-1.

Ex.: Find the number of terms in the expansion of (2x – 3y + 4z)¹⁰⁰

Sol.: Number of terms = ^100+3-1C_3-1 = ¹⁰²C₂

= 102 ! / (2! X 100!)

= (102 x 101) / (2 x 1) = 5151

General term of a multinomial – theorem :

T_r+1 = n! / (n₁! x n₂! ---ⁿm!) x₁ⁿ¹. x₂ⁿ² -----------x_m^nm

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 + ------------ C_nxⁿ -----------------> (i)

Interchanging ‘1’ and ‘x’:

(x+1)ⁿ = C₀ xⁿ + C₁x^n-1 + C₂ x^n-2 + --------- + C_n ------------->(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 C_nxⁿ ---------------> (1)

and (x+1)ⁿ = C₀ xⁿ + C₁ x^n-1 + C₂ x^n-2 + ---------- + C_n -------------> (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 C_n + C₁ C_n-1 + C₂. C_n-2 + ------- + C_n C₀

Then:

Case – 1 : If each term is positive, then

(1+x)ⁿ = C₀ + C₁ x + C₂ x² - -------------- + C_nxⁿ ---------------> (1)

and (1+x)ⁿ = C₀ xⁿ + C₁ x + C₂ x² + ---------- + C_n 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 - -------------- + C_nxⁿ ---------------> (1)

and (1-x)ⁿ = C₀ - C₁ x + C₂ x² + ---------- + (-1)ⁿ C_n 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!) . 2ⁿxⁿ, ‘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.

T_n+1 = ²ⁿC_n 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

= T_n+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) [( x^1/3 / 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 / 2³x 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) x^1/3 + x^-1/5] ⁸

Sol.: General Term T_r+1 = ⁿC_r [(1/2) x^1/3] ^n-r. (x^-1/5)^r

_n-r _-r

= ⁿC_r [(1/2) ^n-r] x ³ x ⁵

Here n = 8

= ⁸C_r (1/2) ^8-r x ^(8-r)/3 ^-r/5

_{40 -8r}

T_r+1 = ⁸C_r (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 : T_r+1 = ¹⁰C_r (-1)^r (x²)^r

Putting r = 5

T₆ = -¹⁰C₅ x¹⁰

Coefficient of x¹⁰ = -¹⁰C₅

In [x – (2/x)] : T_r+1 = ¹⁰C_r (-1)^r (x)^10-r (2/x)^r

= (-1)^r ¹⁰C_r. 2^r. x^10-2r

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 log₁₀^x = z in the given expression :

We have :

( x + x^z)⁵

T₃ = T₂₊₁ = ⁵C₂ (x)^5-2 (x^z)²

= ⁵C₂ x³. x^2z

= 5! / (2! x 3!) x^2z+3

= (5 x 4) / 2! x^2z+3

= T₃ = 10x^2z+3

10,00,000 = 10. x^2z+3

Or x^2z+3 = 10⁵

(10^z)^2z+3 = 10⁵

or 10^2z2+3z = 10⁵

2z² + 3z = 5

[Log₁₀^x = z]

or 2z² + 3z – 5 = 0

or (z-1) (2z+5) = 0

z = 1, - 5 / 2

or log₁₀^x = 1 or log₁₀^x = - 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 + ---- + ^mC_m x^m]

[ⁿC_o – ⁿC₁x + ⁿC₂x² + ------- + (-1)ⁿ ⁿC_nxⁿ]

Coefficient of x = ^mC₁ x ⁿC_o – ^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 : T_2r+1 = ⁴³C_2r. x^2r

Coefficient = ⁴³C_2r

And T_r+2 = ⁴³C_r+1 x^r+1

Coefficient = ⁴³C_r

According to the questions:

⁴³C_2r = ⁴³C_r+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 T_r+1 = ¹⁵C_r. (x²)^15-r. 1/ x^r

T_r+1 = ¹⁵C_r. x^30-3r -------------> (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 5^th and 6^th terms is zero. Find ( a / b) in terms of ‘n’.

Sol.: T₅ = T₄₊₁ = ⁿC₄ (a)^n-4 (-b)⁴

T₅ = ⁿC₄ a^n-4 b⁴

T₆ = T₅₊₁ = ⁿC₅(a)^n-5 (-b)⁵ = -ⁿC₅ a^n-5 b⁵

T₅ + T₆ = 0

ⁿC₄ a^n-4 b⁴ – ⁿC₅ a^n-5 b⁵ = 0

or ⁿC₄ a^n-4b⁴ = ⁿC₅ a^n-5 b⁵

or n!/(4! x n-4!) a^n-4 = n!/(5! x n-5!) a^n-5 b

or a^n-4 / (n-4)(n-5!) = a^n-5 / 5(n-5!) b

or a^n-4 / a^n-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 x^r in the expansion of [x + (1/x)]ⁿ, if it occurs.

Sol.: General term : T_p+1 = ⁿC_p (x)^n-p (1/x) ^p

T_p+1 = ⁿC_p x^n-2p ---------------> (i)

Putting n-2p = r

p = (n - r) / 2

From: (i) T _{(n-r)
/ 2} ₊₁ = ⁿC_{(n-r)
/ 2} x^r

Coefficient of x^r = ⁿC _{(n
– r) / 2}

Q12.: Prove that the coefficient of the term independent of ‘y’ in the expansion of

[(y + 1)/( y ^2/3 – y^1/3 + 1) - (y – 1) / (y – y^1/2)]¹⁰ is 210.

Sol.: We have (y + 1) / (y ^2/3 – y^1/3 + 1)

Putting y = t³, we have

= (t³ + 1³) / (t² – t + 1) = (t + 1) (t² – t + 1) / (t² – t + 1)

= t + 1

(y + 1) / (y^2/3 – y^1/3 + 1) = y^1/3 + 1

and Putting y = a² in (y – 1) / (y – y^1/2 ) :

= (a² – 1) / (a² – a) = (a+1) (a-1) / [a (a-1)]

= (a + 1) / a = 1 + 1 / a

(y – 1) / (y – y^1/2) = 1 + 1 /vy

(y + 1) / (y^2/3 – y^1/3 +1) - (y – 1) / (y – y^1/2)]¹⁰ = [y^1/3 + 1 – 1 – (1/ y^1/2)] ¹⁰

= (y^1/3 – y^-1/2 )¹⁰

In ( y^1/3 – y^-1/2)¹⁰,

T_r+1 = ¹⁰C_r (y^1/3)^10-r. (-y^-1/2)^r

= (-1)^r ¹⁰C_r.^{(10-r)
/ 3 - r/2}

T_r+1 = (-1)^r ¹⁰C_r. 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: x^4-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, T_p+1 = ⁴ⁿC_p(x)^4n-p (1/ x²)^p

T_p+1 = ⁴ⁿC_p x^4n-3p ----------> (i)

Putting :

4n – 3p = 4r

or 4 ( n-r ) / 3 = p

From (i) T_p+1 = ⁴ⁿC_{4(n-r) / 4}. x^4r

Coefficient of x^4r = 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 = T_r+1 = (1+x) [⁴⁰C_r (x³)^r]

= ⁴⁰C_r (1+x) x^3r

= ⁴⁰C_r (x^3r + x^{3r +
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 T_r+1 = ²ⁿC_r(x)^2n-r (1/x) ^r

= ²ⁿC_r. x^2n-2r ---------> (i)

Here 2n – 2r = 0

or n = r

From (i) Tr+1 = ²ⁿC_n

= 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 3^rd, 4^th and 5^th terms in the expansion of (x+a)ⁿ are respectively ‘84’, ‘280’ and ‘560’, find the value of ‘x’, ‘a’ and ‘n’.

Sol.: T_r+1 = ⁿC_r x^n-r. a^r

Putting r = 2, 3 and 4 respectively

T₃ = ⁿC₂ x^n-2. a² = 84 ------------>(i)

T₄ = ⁿC₃ x^n-3 a³ = 280---------->(ii)

and T₅ = ⁿC₄ x^n-4 a⁴ = 560 -------->(iii)

egn (i) x egn(iii) : [ⁿC₂ x^n-2 a²] [ⁿC₄ x^n-4 a⁴] = 84 x 560

= n!/[2! x (n-2)!] x n! / [4! x (n-4)!] . x^2n-6 a⁶ = 84 x 560

or n (n-1) / 2 x n(n-1) (n-2) (n-3) / 4! x x^2n-6 a⁶ = 84 x 560 -------> (iv)

Squaring of egⁿ (ii), we have :

(ⁿC₃ x ^n-3 a³)² = 280²

ⁿC₃ x ⁿC₃ x x^2n-6 x a⁶ = 280²

= n! / [3! x (n-3)!] x n! / [3! x (n-3)!] x x^2n-6 a⁶ = 280²

or n (n-1)(n-2) / 6 x n(n-1) (n-2) (n-3) / 3! x x^2n-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 : T_r+1 = ⁿC_r x^r

2^nd Term : T₂ = ⁿC₁ x

Coefficient = ⁿC₁

3^rd Term : T₃ = ⁿC₂ x²

Coefficient = ⁿC₂

Similarly coefficient of 4^th 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 6^th term in the expansion of [(1/ x^8/3) + x² log₁₀^x]⁸ is 5600. Prove that x =10.

Sol.: T₆ = T₅₊₁ = ⁸C₅ (1/ x^8/3) ^8-5(x² log₁₀^x)⁵

or ⁸C₅ x (1 / x⁸) x c¹⁰ x (log₁₀^x)⁵ = 5600

8! / (5! x 3!) x c² (log₁₀^x)⁵ = 5600

8. 7. 6. / 6 x c² (log₁₀^x)⁵ = 5600

or x²(log₁₀^x)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-----


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