The natural number m, for which the coefficie | vedclass

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Question

$T_{ r +1}={ }^{22} C _{ r }\left( x ^{ m }\right)^{22- r }\left(\frac{1}{ x ^{2}}\right)^{ r }=22 C _{ r } x ^{22 m - mr -2 r }$

$=22 C _{ r } x$

Vedclass · Accepted Answer

$13$

Vedclass · Answer

$T_{ r +1}={ }^{22} C _{ r }\left( x ^{ m }ight)^{22- r }\left(\frac{1}{ x ^{2}}ight)^{ r }=22 C _{ r } x ^{22 m - mr -2 r }$

$=22 C _{ r } x$

$\because 22 C _{3}=22 C _{19}=1540$

$	herefore r =3$ or 19

$Z 2 m - mr -2 r =1$

$m =\frac{2 r +1}{22-5}$

$r =3, \quad m =\frac{7}{19} 
otin N$

$r =19, m =\frac{38+1}{22-19}=\frac{39}{3}=13$

$m =13$

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

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