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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $\left|\frac{y}{x}\right| < 1$ ?

Similar Questions

If $(2-5x)^{-1/5} = a_0 + a_1x + a_2x^2 + \ldots$,then $\frac{a_1}{a_2} = $

If $x$ is so small that $x^2$ and higher powers of $x$ can be neglected,then the approximate value of $\left(1+\frac{3}{4} x\right)^{\frac{1}{2}}\left(1-\frac{2 x}{3}\right)^{-2}$ is

The coefficient of $x^r$ in the expansion of $\frac{1}{\sqrt[3]{(1-2 x)^2}}$ is

If $(a+bx)^{-3} = \frac{1}{27} + \frac{1}{3}x + \dots$,then the ordered pair $(a, b)$ is equal to

If $-\frac{2}{3} < x < \frac{2}{3}$,then the value of the $5^{\text{th}}$ term in the expansion of $\frac{1}{\sqrt[3]{2-3x}}$ when $x=\frac{1}{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module