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

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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 $

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

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 $= \frac{\left(\frac{np+1}{p}ight)\left(\frac{np+1}{p} + 1ight)\left(\frac{np+1}{p} + 2ight)}{3!}$.$= \frac{\left(\frac{np+1}{p}ight)\left(\frac{np+p+1}{p}ight)\left(\frac{np+2p+1}{p}ight)}{3!}$.$= \frac{(np+1)(np+p+1)(np+2p+1)}{p^3 	imes 3!}$.

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

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