The natural number $m,$ for which the coefficient of $x$ in the binomial expansion of ${(x^{m} + \frac{1}{x^{2}})}^{22}$ is $1540,$ is

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Solution

$T_{r + 1} =^{22} C_{r} {(x^{m})}^{22 - r} {(\frac{1}{x^{2}})}^{r}$
$=^{22} C_{r} {(x)}^{22 m - m r - 2 r}$
Given, $^{22} C_{r} = 1540 =^{22} C_{19} =^{22} C_{3}$
$\Rightarrow r = 3, 19$
$∵ 22 m - r m - 2 r = 1$
$\Rightarrow m = \frac{2 r + 1}{22 - r}$
$\Rightarrow r = 3, m = \frac{7}{19} \notin N$
$\Rightarrow r = 19, m = \frac{38 + 1}{22 - 19} = 13 \in N$
Thus, $m = 13$

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