# Solve the following

Question:

If $x^{4}$ occurs in the $r^{\text {th }}$ terms in the expansion of $\left(x^{4}+\frac{1}{x^{3}}\right)^{15}$, then $r=$ ____________

Solution:

for $\left(x^{4}+\frac{1}{x^{3}}\right)^{15}$

n = 15

$T_{r+1}={ }^{15} C_{4}\left(x^{4 r}\right)\left(\frac{1}{x^{3}}\right)^{15-r}$

$={ }^{15} C_{4} x^{4 r} x^{-(45-3 r)}$

$={ }^{15} C_{4} x^{4 r} x^{3 r-45}$

$={ }^{15} C_{4} x^{7 r-45}$

To get x4, 7– 45 = 4

i.e. 7r = 49

i.e r = 7