Question 1

If 26 ( 2^3 3 12 2 + 2^5 5 12 4 + 2^7 7 12 6 + + 2^ 13 13 12 12 ) = 3^ 13 - , then is equal to:

Accepted Answer

Let S = 2^3 3 \ ^ 12 C_ 2 + 2^5 5 \ ^ 12 C_ 4 + 2^7 7 \ ^ 12 C_ 6 + + 2^ 13 13 \ ^ 12 C_ 12 Using the property 1 r+1 \ ^ n C_ r = 1 n+1 \ ^ n+1 C_ r+1 , we can rewrite the terms: S = 2^3 13 \ ^ 13 C_ 3 + 2^5 13 \ ^ 13 C_ 5 + 2^7 13 \ ^ 13 C_ 7 + + 2^ 13 13 \ ^ 13 C_ 13 13S = 2^3 \ ^ 13 C_ 3 + 2^5 \ ^ 13 C_ 5 + 2^7 \ ^ 13 C_ 7 + + 2^ 13 \ ^ 13 C_ 13 We know the binomial expansions: (1+x)^ 13 = \ ^ 13 C_ 0 + \ ^ 13 C_ 1 x + \ ^ 13 C_ 2 x^2 + \ ^ 13 C_ 3 x^3 + + \ ^ 13 C_ 13 x^ 13 (1-x)^ 13 = \ ^ 1

Question 2

If (1 - x^3)^ 10 = _ r=0 ^ 10 a_r x^r (1-x)^ 30-2r , then 9a_9 a_ 10 is equal to __________.

Accepted Answer

The given equation is: (1 - x^3)^ 10 = _ r=0 ^ 10 a_r x^r (1-x)^ 30-2r Using the algebraic identity 1 - x^3 = (1-x)(1+x+x^2), the left hand side can be written as: (1-x)^ 10 (1+x+x^2)^ 10 = _ r=0 ^ 10 a_r x^r (1-x)^ 30-2r Dividing both sides by (1-x)^ 30 : (1-x)^ 10 (1+x+x^2)^ 10 (1-x)^ 30 = _ r=0 ^ 10 a_r x^r (1-x)^ 2r ( 1+x+x^2 (1-x)^2 )^ 10 = _ r=0 ^ 10 a_r ( x (1-x)^2 )^r Let y = x (1-x)^2 . The term inside the bracket on the left hand side can be simplified as: 1+x+x^2 (1-x)^2 = 1-2x+x^2+3x

Question 3

If the coefficients of the middle terms in the binomial expansions of (1 + x)^ 26 and (1 - x)^ 28 , 0, are equal, then the value of is:

Accepted Answer

The expansion of (1 + x)^ 26 has 27 terms. The middle term is T_ 14 . The coefficient of the middle term is ^ 26 C_ 13 ^ 13 . The expansion of (1 - x)^ 28 has 29 terms. The middle term is T_ 15 . The coefficient of the middle term is ^ 28 C_ 14 (- )^ 14 = ^ 28 C_ 14 ^ 14 . Equating the two coefficients: ^ 26 C_ 13 ^ 13 = ^ 28 C_ 14 ^ 14 Since 0, dividing by ^ 13 gives: = ^ 26 C_ 13 ^ 28 C_ 14 Expanding the combinations using factorials: ^ 28 C_ 14 = 28! 14! 14! = 28 27 26! 14 13! 14 13! = 28 27

Question 4

The coefficient of x^2 in the expansion of (2x^2 + 1 x )^ 10 , x 0, is :

Accepted Answer

The general term in the expansion of (2x^2 + 1 x )^ 10 is given by: T_ r+1 = ^ 10 C_ r (2x^2)^ 10-r ( 1 x )^r T_ r+1 = ^ 10 C_ r 2^ 10-r x^ 20-2r x^ -r T_ r+1 = ^ 10 C_ r 2^ 10-r x^ 20-3r For the coefficient of x^2, we equate the power of x to 2: 20 - 3r = 2 3r = 18 r = 6 Substituting r = 6 in the coefficient part, we get: Coefficient = ^ 10 C_ 6 2^ 10-6 = ^ 10 C_ 4 2^4 = 10 9 8 7 4 3 2 1 16 = 210 16 = 3360 Answer: 3360

Question 5

If the sum of the coefficients of x^7 and x^ 14 in the expansion of ( 1 x^3 - x^4 )^n, x 0, is zero, then the value of n is __________.

Accepted Answer

The general term in the expansion of ( 1 x^3 - x^4 )^n is given by: T_ r+1 = ^ n C_ r (x^ -3 )^ n-r (-x^4 )^r = (-1)^r ^ n C_ r x^ 7r - 3n For the coefficient of x^7, we set the exponent to 7: 7r_1 - 3n = 7 r_1 = 3n + 7 7 For the coefficient of x^ 14 , we set the exponent to 14: 7r_2 - 3n = 14 r_2 = 3n + 14 7 Notice that r_2 = r_1 + 1. The sum of the coefficients of x^7 and x^ 14 is zero: (-1)^ r_1 ^ n C_ r_1 + (-1)^ r_2 ^ n C_ r_2 = 0 (-1)^ r_1 ^ n C_ r_1 + (-1)^ r_1 + 1 ^ n C_ r_1 + 1 = 0 (-1)

Question 6

In the expansion of (9x- 1 3 x )^ 18 , x>0, if the term independent of x is (221)k, then k is equal to:

Accepted Answer

The general term in the expansion of (9x - 1 3 x )^ 18 is given by: T_ r+1 = ^ 18 C_ r (9x)^ 18-r (- 1 3 x )^r T_ r+1 = ^ 18 C_ r 9^ 18-r (- 1 3 )^r x^ 18-r x^ -r/2 T_ r+1 = ^ 18 C_ r 9^ 18-r (- 1 3 )^r x^ 18 - 3r 2 For the term independent of x, the exponent of x must be zero: 18 - 3r 2 = 0 3r 2 = 18 r = 12 Substituting r = 12 into the general term: T_ 13 = ^ 18 C_ 12 9^ 18-12 (- 1 3 )^ 12 T_ 13 = ^ 18 C_ 12 9^6 ( 1 3 )^ 12 T_ 13 = ^ 18 C_ 12 (3^2)^6 1 3^ 12 T_ 13 = ^ 18 C_ 12 3^ 12 1 3^ 12 = ^

Question 7

Let the smallest value of k N , for which the coefficient of x^3 in (1+x)^3 + (1+x)^4 + (1+x)^5 + + (1+x)^ 99 + (1+kx)^ 100 , x 0, is (43n + 101 4 ) (^ 100 C_3 ) for some n N , be p. Then the value of

Accepted Answer

The coefficient of x^3 in the given expression is the sum of the coefficients of x^3 in each term. The coefficient of x^3 in _ r=3 ^ 99 (1+x)^r + (1+kx)^ 100 is: _ r=3 ^ 99 ^ r C_3 + k^3 ^ 100 C_3 Using the identity _ r=k ^ n ^ r C_k = ^ n+1 C_ k+1 , we get: _ r=3 ^ 99 ^ r C_3 = ^ 100 C_4 Thus, the total coefficient of x^3 is ^ 100 C_4 + k^3 ^ 100 C_3. Given that this coefficient is equal to (43n + 101 4 ) ^ 100 C_3, we have: ^ 100 C_4 + k^3 ^ 100 C_3 = (43n + 101 4 ) ^ 100 C_3 Dividing both sid

Question 8

If for 3 r 30, 30 30-r + 3 30 31-r + 3 30 32-r + 30 33-r = m r , then m equals:

Accepted Answer

Using the property ^ n C_ r = ^ n C_ n-r , the given expression can be written as: ^ 30 C_ r + 3 ^ 30 C_ r-1 + 3 ^ 30 C_ r-2 + ^ 30 C_ r-3 This expression represents the coefficient of x^r in the expansion of: (1+x)^ 30 + 3x(1+x)^ 30 + 3x^2(1+x)^ 30 + x^3(1+x)^ 30 Taking (1+x)^ 30 common, we get: (1+x)^ 30 (1 + 3x + 3x^2 + x^3) = (1+x)^ 30 (1+x)^3 = (1+x)^ 33 The coefficient of x^r in (1+x)^ 33 is ^ 33 C_ r . Comparing this with ^ m C_ r , we get m = 33. Answer: 33

Question 9

The number of elements in the set S = \ (r, k) : k Z and ^ 36 C_ r+1 = 6 ( ^ 35 C_r ) (k^2 - 3) \ , is :

Accepted Answer

Given equation: ^ 36 C_ r+1 = 6 (^ 35 C_r ) k^2 - 3 Using the property ^ n C_r = n r ^ n-1 C_ r-1 , we can write: ^ 36 C_ r+1 = 36 r+1 ^ 35 C_r Substituting this into the given equation: 36 r+1 ^ 35 C_r = 6 k^2 - 3 ^ 35 C_r For ^ 35 C_r to be defined, r must be an integer such that 0 r 35. Thus, ^ 35 C_r 0. Dividing both sides by ^ 35 C_r: 36 r+1 = 6 k^2 - 3 6 r+1 = 1 k^2 - 3 k^2 - 3 = r+1 6 k^2 = 3 + r+1 6 For k Z , k^2 must be a perfect square integer. This requires r+1 to be a multiple of 6.

Question 10

Given below are two statements : Statement I: 25^ 13 +20^ 13 +8^ 13 +3^ 13 is divisible by 7. Statement II: The integral part of (7+4 3 )^ 25 is an odd number. In the light of the above statements, ch

Accepted Answer

Statement I: 25^ 13 + 3^ 13 → divisible by (25 + 3) 20^ 13 + 8^ 13 → divisible by (20 + 8) divisible by 7. TRUE Statement II: For (7+4 3 )^ 25 , let = 7+4 3 and = 7-4 3 (conjugate). Then = 1 and + = 14. Since 0 The fractional part of ^ 25 is 1- ^ 25 , making the integral part odd. TRUE. Both Statement I and Statement II are true.

Question 11

The sum of the coefficients of x^ 499 and x^ 500 in (1+x)^ 1000 +x(1+x)^ 999 +x^ 2 (1+x)^ 998 + +x^ 1000 is :

Accepted Answer

S = (1+x)^ 1000 + x(1+x)^ 999 + x^2(1+x)^ 998 + + x^ 1000 This is a geometric series with first term (1+x)^ 1000 , common ratio x 1+x , and 1001 terms. S = (1+x)^ 1000 1 - ( x 1+x )^ 1001 1 - x 1+x = (1+x)^ 1001 - x^ 1001 Required sum = coefficient of x^ 499 + coefficient of x^ 500 in (1+x)^ 1001 - x^ 1001 = ^ 1001 C_ 499 + ^ 1001 C_ 500 = ^ 1002 C_ 500

Question 12

Let S = 1 25! + 1 3!23! + 1 5!21! + up to 13 terms. If 13 ~S = 2^ k n! , k ~N , then n+k is equal to

Accepted Answer

The general term is 1 (2r-1)!(27-2r)! for r=1 to 13. Multiply and divide by 26!: 1 (2r-1)!(27-2r)! = 1 26! 26 2r-1 S = 1 26! _ r=1 ^ 13 26 2r-1 = 1 26! [ 26 1 + 26 3 + + 26 25 ] Sum of odd-indexed binomial coefficients: 26 odd = 2^ 25 S = 2^ 25 26! 13S = 13 2^ 25 26! = 13 2^ 25 26 25! = 2^ 25 2 25! = 2^ 24 25! So k = 24,\; n = 25, giving n + k = 49.

Question 13

The value of ^ 100 C _ 50 51 + ^ 100 C _ 51 52 + .+ ^ 100 C _ 100 101 is :

Accepted Answer

Using n r r+1 = 1 n+1 n+1 r+1 : _ r=50 ^ 100 100 r r+1 = 1 101 _ r=50 ^ 100 101 r+1 = 1 101 _ k=51 ^ 101 101 k . Since 101 is odd, by symmetry _ k=51 ^ 101 101 k = _ k=0 ^ 50 101 k = 2^ 101 2 = 2^ 100 . Required value = 2^ 100 101 .

Question 14

The sum of all possible values of n N , so that the coefficients of x, x^ 2 and x^ 3 in the expansion of (1+x^ 2 )^ 2 (1+x)^ n , are in arithmetic progression is :

Accepted Answer

(1+x^2)^2 = 1 + 2x^2 + x^4. Multiply with (1+x)^n: Coefficient of x: n. Coefficient of x^2: n 2 + 2 = n(n-1) 2 + 2. Coefficient of x^3: n 3 + 2n = n(n-1)(n-2) 6 + 2n. For AP: 2 ( n(n-1) 2 +2 ) = n + n(n-1)(n-2) 6 + 2n. Simplifying: n^3 - 9n^2 + 26n - 24 = 0. (n-2)(n-3)(n-4) = 0 n = 2, 3, 4. Sum = 2 + 3 + 4 = 9.

Question 15

Let C _ r denote the coefficient of x^ r in the binomial expansion of (1+x)^ n , n N , 0 r n . If P_ n =C_ 0 -C_ 1 + 2^ 2 3 C_ 2 - 2^ 3 4 C_ 3 + . .+ (-2)^ n n+1 C_ n , then the value of _ n=1 ^ 25 1

Accepted Answer

Using the identity 1 r+1 n r = 1 n+1 n+1 r+1 , we get P_n = _ r=0 ^ n (-2)^r r+1 C_r = 1 n+1 _ s=1 ^ n+1 (-2)^ s n+1 s . Since (1-2)^ n+1 = (-1)^ n+1 = _ s=0 ^ n+1 (-2)^s n+1 s , we have _ s=1 ^ n+1 (-2)^s n+1 s = (-1)^ n+1 - 1. For even n = 2k: P_ 2k = 1 - (-1) 2(2k+1) = 1 2k+1 . Thus 1 P_ 2n = 2n+1, and _ n=1 ^ 25 (2n+1) = 2 25 26 2 + 25 = 675.

Question 16

The coefficient of x^ 48 in (1+x)+2(1+x)^ 2 +3(1+x)^ 3 + +100(1+x)^ 100 is equal to

Accepted Answer

Let 1 + x = r S = 1 r + 2 r^2 + 3 r^3 + + 100r^ 100 ......(1) (AGP) rS = 1 r^2 + 2 r^3 + + 99r^ 100 + 100r^ 101 ......(2) (1) - (2) gives S = - (1+x)^ 101 x^2 + 1 x^2 + 100(1+x)^ 101 x coefficient x^ 48 in S = - coefficient of x^ 48 in (1+x)^ 101 x^2 + 100 Coefficient of x^ 48 in (1+x)^ 101 x = 100 ^ 101 C_ 49 - ^ 101 C_ 50

Question 17

If ( 1 ^ 15 C _ 0 + 1 ^ 15 C _ 1 ) ( 1 ^ 15 C _ 1 + 1 ^ 15 C _ 2 ) ( 1 ^ 15 C _ 12 + 1 ^ 15 C _ 13 )= ^ 13 ^ 14 C _ 0 ^ 14 C _ 1 ^ 14 C _ 12 , then 30 is equal to \_\_\_\_.

Accepted Answer

The general term of the product is T_r = 1 ^ 15 C_ r-1 + 1 ^ 15 C_r for r = 1, 2, , 13. Using the formula ^ n C_r = n r ^ n-1 C_ r-1 , we have ^ 15 C_r = 15 r ^ 14 C_ r-1 and ^ 15 C_ r-1 = 15 15-(r-1) ^ 14 C_ r-1 = 15 16-r ^ 14 C_ r-1 . Substituting these into T_r: T_r = 16-r 15 ^ 14 C_ r-1 + r 15 ^ 14 C_ r-1 = 16-r+r 15 ^ 14 C_ r-1 = 16 15 ^ 14 C_ r-1 . The product is _ r=1 ^ 13 T_r= _ r=1 ^ 13 16 15 ^ 14 C_ r-1 = 16^ 13 15^ 13 _ r=1 ^ 13 ^ 14 C_ r-1 . The denominator of the product is 15^ 13 (

Question 18

If the coefficient of x in the expansion of (a x^ 2 +b x+c )(1-2 x)^ 26 is -56 and the coefficients of x^ 2 and x^ 3 are both zero, then a + b + c is equal to :

Accepted Answer

In (1-2x)^ 26 : T_0 = 1, T_1 = -52, T_2 = 1300, T_3 = -20800. Coefficient of x: b - 52c = -56 ...(1) Coefficient of x^2: a - 52b + 1300c = 0 ...(2) Coefficient of x^3: -52a + 1300b - 20800c = 0 ...(3) From (1): b = 52c - 56. Substituting in (2): a = 1404c - 2912. Substituting in (3): -26208c + 78624 = 0 c = 3. Thus b = 100, a = 1300. a + b + c = 1300 + 100 + 3 = 1403.

Binomial Theorem — Previous Year Questions & Solutions