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

Question

Aicharya · Accepted Answer

Solution: Suppose the (r + 1) th term in the given expansion contains the coefficient of x -17 . Then, we have $\text{T}_{\text{r}+1}={^\text{15}}\text{C}_{\text{r}}(\text{x}^{4})^{15-\text{r}}\Big(\frac{-1}{\text{x}^{3}}\Big)^{\text{r}}$ $\Rightarrow (1)^{\text{r}}\ {^\text{15}}\text{C}_{\text{r}-1}\ \text{x}^{60-4\text{r}-3\text{r}}$ For this term to contain x -17 , we mst have $60-7\text{r}=-17$ $\Rightarrow 7\text{r}=77$ $\Rightarrow \text{r}=11$ $\therefore$ Required coefficient $=(-1)^{11}\ {^\text{15}}\text{C}_{\text{11}}=-\frac{15\times14\times13\times12}{4\times3\times2}=-1365$

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

Solution