The {(r + 1)^{th}} term in the expansion of { | vedclass

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

Question

(B) The general term for the expansion of $(1 - x)^{-n}$ is given by $T_{r+1} = \binom{n+r-1}{r} x^r$.For the expansion of $(1 - x)^{-4}$,we have $n =

Vedclass · Accepted Answer

$\frac{(r + 1)(r + 2)(r + 3)}{6}x^r$

Vedclass · Answer

(B) The general term for the expansion of $(1 - x)^{-n}$ is given by $T_{r+1} = \binom{n+r-1}{r} x^r$.For the expansion of $(1 - x)^{-4}$,we have $n = 4$.Thus,$T_{r+1} = \binom{4+r-1}{r} x^r = \binom{r+3}{r} x^r$.Using the property $\binom{n}{r} = \binom{n}{n-r}$,we get $\binom{r+3}{r} = \binom{r+3}{3}$.Expanding the binomial coefficient: $\binom{r+3}{3} = \frac{(r+3)(r+2)(r+1)}{3 	imes 2 	imes 1} = \frac{(r+1)(r+2)(r+3)}{6}$.Therefore,$T_{r+1} = \frac{(r+1)(r+2)(r+3)}{6} x^r$.

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 ${(r + 1)^{th}}$ term in the expansion of ${(1 - x)^{-4}}$ is:

Similar Questions

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

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

If the third term in the binomial expansion of $(1 + x)^m$ is $-\frac{1}{8}x^2$,then the rational value of $m$ is

If $|x| < 1$,then the number of terms in the expansion of $[\frac{1}{2}(1 \cdot 2 + 2 \cdot 3 x + 3 \cdot 4 x^2 + . . . . . . \infty)]^{-25}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module