Question

The coefficient of $x^{-17}$ in the expansion of ${(x^4-\frac{1}{x^3})}^{15}$ is

• A. 1365
• B. -1365
• C. 3003
• D. -3003

Answer 1

The correct option is B

-1365

-1365

Suppose the (r+1)th term in the given expansion contains the coefficient of $x^{-17}$

Then, we have:

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

$=(-1{)}^r{}^{15}{C}_r{x}^{60-4r-3r}$

For this term to contain $x^{-17}$,

we must have: 60-7r=-17

$\Rightarrow 7r=77$

$\Rightarrow r=11$

$\therefore$ Required coefficient $=(-1{)}^{11}{}^{15}{C}_{11}$

$=\frac{15\times 14\times 13\times 12}{4\times 3\times 2}=-1365$