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

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