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

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(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:

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