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

Question

In the expansion of ((1/x2)-x3)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

((1/x2)-x3)n Tr+1=(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 According to question (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 (x21​−x3)nn∈N, if the sum of the coefficients of x5 and x10 is 0, then n is:

25

20

15

None of these

(x21​−x3)n Tr+1​=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 According to question (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 $(\frac{1}{x ^{2}} - x^{3})^{n} n \in N,$ if the sum of the coefficients of $x^{5}$ and $x^{10}$ is $0,$ then $n$ is: