First slide

Binomial theorem for positive integral Index

Question

The third term in the expansion of ${(x^{2} - \frac{1}{x^{3}})}^{n}$ is independent of $x$ , when $n$ is equal to

Easy

A

2
B

3
C

4
D

5

Solution

Given expansion ${(x^{2} - \frac{1}{x^{3}})}^{n}$ is

We have general term in the expansion ${(x + a)}^{n}$

$(∴ T_{r + 1} =^{n} C_{r} x^{n - r} {(a)}^{r}$ be the expansion of ${(x + a)}^{n})$

$T_{2 + 1} =^{n} C_{2} {(x^{2})}^{n - 2} {(- \frac{1}{x^{3}})}^{2}$

$T_{2 + 1} =^{n} C_{2} {(x^{2})}^{n - 2} {(- 1)}^{2} {(x^{- 3})}^{2}$

$T_{3} =^{n} C_{2} x^{2 n - 10}$

$x^{2 n - 10}$ compare with $x^{0}$ for finding $r$

$\Rightarrow x^{2 n - 10} = x^{0}$

will contain $x$ if $2 n - 10$ $= 0$

$∵ n = 5$ .

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