In the expansion of x3-1x2n- n - N- if the sum of the coefficients of x5 and x10 is 0- then n is

(x^3 1x^2)^n General term T_r+1=n!r!(n r)!( 1)^n rx^5r 2n nbsp; If 5r 2n=5, then 5r=2n+5 or r=2n5+1 If 5r 2n=10, then 5r=2n+10 or r=2n5+2 x^5 and x^10 terms ...

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Byju's Answer

Standard XIII

Mathematics

General Term of Binomial Expansion

In the expans...

Question

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

25

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15

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Solution

The correct option is C $15$
${(x^{3} - \frac{1}{x^{2}})}^{n}$
General term $T_{r + 1} = \frac{n!}{r! (n - r)!} (- 1)^{n - r} x^{5 r - 2 n}$
If $5 r - 2 n = 5$ , then $5 r = 2 n + 5$ or $r = \frac{2 n}{5} + 1$

If $5 r - 2 n = 10$ , then $5 r = 2 n + 10$ or $r = \frac{2 n}{5} + 2$
$x^{5}$ and $x^{10}$ terms occur if $n = 5 k$

Given, the sum of coefficients $x^{5}$ and $x^{10}$ is zero.
$\Rightarrow \frac{(5 k)!}{(2 k + 1)! (3 k - 1)!} - \frac{(5 k)!}{(2 k + 2)! (3 k - 2)!} = 0$
or, $\frac{1}{3 k - 1} - \frac{1}{2 k + 2} = 0$
or, $k = 3 \Rightarrow n = 15$

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

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