What is the coefficient of $\frac{y^3}{x^8}$ in $(x+y)^{-5}$, when $\left|\frac{y}{x}\right|

Question

What is the coefficient of $\frac{y^3}{x^8}$ in $(x+y)^{-5}$, when $\left|\frac{y}{x}\right| < 1$ ?

Aicharya · Accepted Answer

Since $(x+y)^{-5}=\frac{1}{(x+y)^5}=\frac{1}{x^5}\left(1+\frac{y}{x}\right)^{-5},\left|\frac{y}{x}\right| < 1$ $\because$ Coefficient of $\frac{y^3}{x^8} \operatorname{in}(x+y)^{-5}$ $\begin{aligned} & =\text { coefficient of } \frac{y^3}{x^8} \text { in } \frac{1}{x^5}\left(1+\frac{y}{x}\right)^{-5} \ & =\text { coefficient of } \frac{y^3}{x^3} \operatorname{in}\left(1+\frac{y}{x}\right)^{-5} \ & =\frac{-5(-5-1)(-5-2)}{3 !}=-35 \end{aligned}$ Hence, option (a) is correct.

What is the coefficient of \(\frac{y^3}{x^8}\) in \((x+y)^{-5}\), when \(\left|\frac{y}{x}\right| < 1\) ?

Solution