First slide

Binomial theorem for positive integral Index

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

Moderate

A

25
B

20
C

15
D

None of these

Solution

${(x^{3} - \frac{1}{r^{2}})}^{n}$
General term $T_{r + 1} = \frac{n!}{r! (n - r)!} (- 1)^{n - r} x^{5 n - 2 n}$
If $5 r - 2 n = 5, then 5 r = 2 n + 5 or r = \frac{2 n}{5} + 1$
If $5 r - 2 n = 10, then 5 r = 2 n + 10 or r = \frac{2 n}{5} + 2$
$x^{5} and x^{10} terms occurs if n = 5 k$
Given that sum of x⁵ and x¹⁰ 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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