What is the coefficient of frac{y^3}{x^8} in | vedclass

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(A) Given the expression $(x+y)^{-5} = \frac{1}{x^5} \left(1 + \frac{y}{x}\right)^{-5}$ for $\left|\frac{y}{x}\right| < 1$. Using the binomial expansi

Vedclass · Accepted Answer

-$35$

Vedclass · Answer

(A) Given the expression $(x+y)^{-5} = \frac{1}{x^5} \left(1 + \frac{y}{x}ight)^{-5}$ for $\left|\frac{y}{x}ight| Using the binomial expansion $(1+z)^{-n} = 1 - nz + \frac{n(n+1)}{2!}z^2 - \frac{n(n+1)(n+2)}{3!}z^3 + \dots$,where $z = \frac{y}{x}$ and $n = 5$. The general term is given by $\binom{-5}{r} \left(\frac{y}{x}ight)^r$. We want the coefficient of $\frac{y^3}{x^8} = \frac{1}{x^5} \cdot \left(\frac{y}{x}ight)^3$. This corresponds to the term where $r = 3$ in the expansion of $\left(1 + \frac{y}{x}ight)^{-5}$. The coefficient is $\binom{-5}{3} = \frac{(-5)(-6)(-7)}{3 	imes 2 	imes 1} = \frac{-210}{6} = -35$. Thus,the coefficient is $-35$. Hence,option $A$ is correct.

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

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