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7.Binomial Theorem
hard
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
A
$19$
B
$3$
C
$13$
D
$18$
(JEE MAIN-2020)
Solution
$T_{ r +1}={ }^{22} C _{ r }\left( x ^{ m }\right)^{22- r }\left(\frac{1}{ x ^{2}}\right)^{ r }=22 C _{ r } x ^{22 m – mr -2 r }$
$=22 C _{ r } x$
$\because 22 C _{3}=22 C _{19}=1540$
$\therefore r =3$ or 19
$Z 2 m – mr -2 r =1$
$m =\frac{2 r +1}{22-5}$
$r =3, \quad m =\frac{7}{19} \notin N$
$r =19, m =\frac{38+1}{22-19}=\frac{39}{3}=13$
$m =13$
Standard 11
Mathematics
