The natural number $m$, for which the coefficient of $x$ in the binomial expansion of $\left(x^{m}+\

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 ,

Professor PC Thomas Classes · Answer

(13) : We have, $\left(x^{m}+\frac{1}{x^{2}}ight)^{22}$\ $T_{r+1}={ }^{22} C_{r}\left(x^{m}ight)^{22-r}\left(\frac{1}{x^{2}}ight)^{r}$\ $={ }^{22} C_{r} x^{22 m-m r-2 r}$\ Since, ${ }^{22} C_{3}={ }^{22} C_{19}=1540 \quad 	herefore \quad r=3$ or 19\ Now, for coeff. of $x, 22 m-m r-2 r=1$\ $\Rightarrow m=\frac{2 r+1}{22-r}$\ At $r=3, m=\frac{7}{19} 
otin N$\ At $r=19, m=\frac{39}{3}=13 \in N \quad 	herefore m=13$\