The natural number m,


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__________.


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

$T_{r+1}={ }^{22} C_{r} \cdot x^{22 m-m r-2 r}$

$\because 22 m-m r-2 r=1$

$\Rightarrow r=\frac{22 m-1}{m+2} \Rightarrow r=22-\frac{3 \cdot 3 \cdot 5}{m+2}$

So, possible value of $m=1,3,7,13,43$

But ${ }^{22} C_{r}=1540$

$\therefore$ Only possible value of $m=13$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now