The first four terms in the expansion of (1 - | vedclass

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(C) The binomial expansion for $(1 + z)^n$ is given by $1 + nz + \frac{n(n-1)}{2!}z^2 + \frac{n(n-1)(n-2)}{3!}z^3 + \dots$Here,$n = \frac{3}{2}$ and $

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$1 - \frac{3}{2}x + \frac{3}{8}x^2 + \frac{x^3}{16}$

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(C) The binomial expansion for $(1 + z)^n$ is given by $1 + nz + \frac{n(n-1)}{2!}z^2 + \frac{n(n-1)(n-2)}{3!}z^3 + \dots$Here,$n = \frac{3}{2}$ and $z = -x$.Substituting these values:Term $1$: $1$Term $2$: $\frac{3}{2}(-x) = -\frac{3}{2}x$Term $3$: $\frac{\frac{3}{2}(\frac{3}{2}-1)}{2!}(-x)^2 = \frac{\frac{3}{2} \cdot \frac{1}{2}}{2}x^2 = \frac{3}{8}x^2$Term $4$: $\frac{\frac{3}{2}(\frac{3}{2}-1)(\frac{3}{2}-2)}{3!}(-x)^3 = \frac{\frac{3}{2} \cdot \frac{1}{2} \cdot (-\frac{1}{2})}{6}(-x)^3 = \frac{-\frac{3}{8}}{6}(-x^3) = \frac{3}{48}x^3 = \frac{1}{16}x^3$Thus,the first four terms are $1 - \frac{3}{2}x + \frac{3}{8}x^2 + \frac{1}{16}x^3$.

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

The first four terms in the expansion of $(1 - x)^{3/2}$ are

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