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,$T_{r+1} = \frac{(r+1)(r+2)(r+3)}{6} x^r$.

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

Similar Questions

If $|x| < 1$,then the value of $1 + n\left( \frac{2x}{1 + x} \right) + \frac{n(n + 1)}{2!}\left( \frac{2x}{1 + x} \right)^2 + \dots \infty$ will be

The coefficient of $x^2$ in the expansion of $(1+x)^2(8-x)^{-\frac{1}{3}}$ is

$\frac{1}{4}-\frac{5}{4 \cdot 8}+\frac{5 \cdot 7}{4 \cdot 8 \cdot 12}-\ldots=$

If $y = \frac{3}{4} + \frac{3 \times 5}{4 \times 8} + \frac{3 \times 5 \times 7}{4 \times 8 \times 12} + \ldots$ to $\infty$,then

If $y = 3x + 6x^2 + 10x^3 + \dots$,then the value of $x$ in terms of $y$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module