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, 7r – 45 = 4
i.e. 7r = 49
i.e r = 7