First slide

Binomial theorem for positive integral Index

Question

If $x^{2 k}$ occurs in the expansion of ${(x + \frac{1}{x^{2}})}^{n - 3},$ then

Moderate

A

$n - 2 k$ is a multiple of $2$
B

$n - 2 k$ is a multiple of $3$
C

$k = 0$
D

none of these

Solution

$T_{r + 1}$ the $(r + 1) th$ term in the expansion of ${(x + \frac{1}{x^{2}})}^{n - 3}$ is given by
$T_{r + 1} =^{n - 3} C_{r} (x)^{n - 3 - r} {(\frac{1}{x^{2}})}^{r} =^{n - 3} C_{r} x^{n - 3 - 3 r}$
As $x^{2 k}$ occurs in the expansion of ${(x + \frac{1}{x^{2}})}^{n - 3},$ we must
haven $n - 3 - 3 r = 2 k$ for some non-negative integer $r .$
$\Rightarrow 3 (1 + r) = n - 2 k \Rightarrow n - 2 k$ is a multiple of $3 .$

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