Question

Find the coefficients of $\rm x^{40}$ in the expansion of $\rm \Big(\frac{1}{x^{2}} + x^{4} \Big)^{18}$.

Answer 1

Solution :

$\rm \Big(\frac{1}{x^{2}} + x^{4} \Big)^{18}$.

We know that ,
*****************************************
If ( x + a )ⁿ ,

$t_{r+1} = ^nC_{r}x^{n-r}a^{r}$

********************************************

Here ,

$\rm\t_{r+1} = ^{18}C_{r} \(frac{1}{x^{2}})^{18-r}\times(x^{4})^{r}$

= $^{18}C_{r}x^{-2(18-r)}x^{4r}$

= $^{18}C_{r}x^{-36+2r+4r}$

But according to the problem given,

-36+2r+4r = 40

=> 6r = 40 + 36

=> 6r = 76

=>r = 76/6

=> r = 38/3 ( fraction )

Therefore ,

In the expansion we doesn't get x^40 term.