Question

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

Answer 1

Answer: 13

$T_{r+1}={}^{22}{C}_r{\left(x^m\right)}^{22-r}{\left(\frac{1}{x^2}\right)}^r={}^{22}{C}_r{x}^{22m-mr-2r}$
$={}^{22}{C}_{r}x$
$\because {}^{22}{C}_3={}^{22}{C}_{19}=1540$
$\therefore r=3$ or 19
$22m-mr-2r=1$
$m=\frac{2r+1}{22-5}$
$r=3,m=\frac{7}{19}\notin N$
$r=19,m=\frac{38+1}{22-19}=\frac{39}{3}=13$
$m=13$