For n, p in N-{1},the coefficient of x^3 in f | vedclass

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

Question

(A) The given expression is $\frac{(1-x)^{-1/p}}{(1-x)^n} = (1-x)^{-(n + 1/p)} = (1-x)^{-\frac{np+1}{p}}$.Using the binomial expansion for any index $

Vedclass · Accepted Answer

$\frac{(n p+1)(n p+p+1)(n p+2 p+1)}{p^3 	imes 3!}$

Vedclass · Answer

(A) The given expression is $\frac{(1-x)^{-1/p}}{(1-x)^n} = (1-x)^{-(n + 1/p)} = (1-x)^{-\frac{np+1}{p}}$.Using the binomial expansion for any index $(1-x)^{-k} = 1 + kx + \frac{k(k+1)}{2!}x^2 + \frac{k(k+1)(k+2)}{3!}x^3 + \dots$,where $k = \frac{np+1}{p}$.The coefficient of $x^3$ is $\frac{k(k+1)(k+2)}{3!}$.Substituting $k = \frac{np+1}{p}$:Coefficient $= \frac{\left(\frac{np+1}{p}ight)\left(\frac{np+1}{p} + 1ight)\left(\frac{np+1}{p} + 2ight)}{3!}$.$= \frac{\left(\frac{np+1}{p}ight)\left(\frac{np+p+1}{p}ight)\left(\frac{np+2p+1}{p}ight)}{3!}$.$= \frac{(np+1)(np+p+1)(np+2p+1)}{p^3 	imes 3!}$.

List-$I$	List-$II$
$(A)$ $(1-x)^{-n}$	$(i)$ $\frac{x}{x+1}$
$(B)$ $(1+x)^{-n}$	$(ii)$ $1-nx+\frac{n(n+1)}{2!}x^2-\dots$ if $\|x\| < 1$
$(C)$ If $x>1$,then $1+\frac{1}{x}+\frac{1}{x^2}+\dots$ is	$(iii)$ $1+nx+\frac{n(n+1)}{2!}x^2+\dots$ if $\|x\| < 1$
$(D)$ If $\|x\|>1$,then $1-\frac{2}{x^2}+\frac{3}{x^4}-\frac{4}{x^6}+\dots$ is	$(iv)$ $\frac{x}{x-1}$
	$(v)$ $\frac{x^4}{(x^2+1)^2}$
	$(vi)$ $\frac{x^4}{(x^2-1)^2}$

For $n, p \in N-\{1\}$,the coefficient of $x^3$ in $\frac{(1-x)^{-1 / p}}{(1-x)^n}$ is:

Similar Questions

The sum of the coefficients of $x^{-3/2}$ and $x^3$ in the expansion of $\sqrt{3+x} + \sqrt{5+x}$ when $3 < x < 5$ is

The sum of the series $\frac{3}{4 \cdot 8} - \frac{3 \cdot 5}{4 \cdot 8 \cdot 12} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16} - \dots$ is:

For $x=\frac{5}{7}$,if $t_k$ is the first negative term in the expansion of $(1+x)^{7/5}$,then $t_1+t_2+\ldots+t_k=$

To expand $(1 + 2x)^{-1/2}$ as an infinite series,the range of $x$ should be

Vedclass Test Series

Exam Paper Generator

Online Exam Module