In the expansion of (x3-(1/x2))n, n ∈ N, if the sum of the coefficients of x5 and x10 is 0 , then n is :

Question

In the expansion of (x3-(1/x2))n, n ∈ N, if the sum of the coefficients of x5 and x10 is 0 , then n is : (A) 25 (B) 20 (C) 15 (D) None of these

Tardigrade Admin · Accepted Answer

(x3-(1/x2))n General term =(n !/r !(n-r) !)(-1)n-r x5 r-2 n If 5 r-2 n=5, then 5 r=2 n+5 ⇒ r=(2 n/5)+1 If 5 r-2 n=10, then 5 r=2 n+10 ⇒ r=(2 n/5)+2 Let n=5 k Now (5 k !/(2 k+1) !(3 k-1) !)-(5 k !/(2 k+2) !(3 k-2) !)=0 ⇒ (1/3 k -1)-(1/2 k +2)=0 ⇒ k =3 ⇒ n =15

Q. In the expansion of (x3−x21​)n,n∈N, if the sum of the coefficients of x5 and x10 is 0 , then n is :

25

20

15

None of these

(x3−x21​)n General term =r!(n−r)!n!​(−1)n−rx5r−2n If 5r−2n=5, then 5r=2n+5⇒r=52n​+1 If 5r−2n=10, then 5r=2n+10⇒r=52n​+2 Let n=5k Now (2k+1)!(3k−1)!5k!​−(2k+2)!(3k−2)!5k!​=0 ⇒3k−11​−2k+21​=0 ⇒k=3⇒n=15

Q. In the expansion of $(x^{3} - \frac{1}{x ^{2}})^{n}, n \in N$ , if the sum of the coefficients of $x^{5}$ and $x^{10}$ is 0 , then $n$ is :