The coefficient of {x^r} in the expansion of | vedclass

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(B) The general term in the expansion of $(1+x)^n$ is given by $\binom{n}{r} x^r = \frac{n(n-1)...(n-r+1)}{r!} x^r$. For $(1-2x)^{-1/2}$,we have $n =

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$\frac{(2r)!}{2^r(r!)^2}$

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(B) The general term in the expansion of $(1+x)^n$ is given by $\binom{n}{r} x^r = \frac{n(n-1)...(n-r+1)}{r!} x^r$. For $(1-2x)^{-1/2}$,we have $n = -1/2$ and the term is $\binom{-1/2}{r} (-2x)^r$. The coefficient of ${x^r}$ is $\binom{-1/2}{r} (-2)^r$. $= \frac{(-1/2)(-3/2)(-5/2)...(-(2r-1)/2)}{r!} (-2)^r$. $= \frac{(-1)^r [1 	imes 3 	imes 5 	imes ... 	imes (2r-1)]}{2^r r!} (-2)^r$. $= \frac{(-1)^r [1 	imes 3 	imes 5 	imes ... 	imes (2r-1)]}{2^r r!} (-1)^r 2^r$. $= \frac{1 	imes 3 	imes 5 	imes ... 	imes (2r-1)}{r!}$. Multiplying numerator and denominator by $2 	imes 4 	imes 6 	imes ... 	imes (2r) = 2^r r!$,we get: $= \frac{(1 	imes 3 	imes ... 	imes (2r-1)) 	imes (2 	imes 4 	imes ... 	imes 2r)}{r! 	imes 2^r r!} = \frac{(2r)!}{2^r (r!)^2}$.

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	$(iv)$ $\frac{x}{x-1}$
	$(v)$ $\frac{x^4}{(x^2+1)^2}$
	$(vi)$ $\frac{x^4}{(x^2-1)^2}$

The coefficient of ${x^r}$ in the expansion of ${(1 - 2x)^{-1/2}}$ is

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