Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

# Find the coefficient of

Question:

Find the coefficient of $1 / x^{17}$ in the expansion of $\left(x^{4}-\frac{1}{x^{3}}\right)^{15}$

Solution:

Given $\left(x^{4}-\frac{1}{x^{3}}\right)^{15}$

From the standard formula of $T_{r+1}$ we can write given expression as

$T_{r+1}={ }^{15} C_{r}\left(x^{4}\right)^{15-r}\left(-\frac{1}{x^{3}}\right)^{r}={ }^{15} C_{r} x^{60-4 r}(-1)^{r} x^{-3 r}={ }^{15} C_{r} x^{60-7 r}(-1)^{r}$

For the coefficient $\mathrm{x}^{-17}$, we have

$60-7 r=-17$

Therefore, $r=11$

Then above expression becomes

$T_{11+1}={ }^{15} C_{11} x^{60-77}(-1)^{11}$

Coefficient of $x^{-17}=\frac{-15 \times 14 \times 13 \times 12 \times 11 !}{11 ! \times 4 \times 3 \times 2 \times 1}$

$=-15 \times 7 \times 13=-1365$