A) \[\frac{{{x}^{r}}}{r!}\]
B) \[\frac{(r+1)(r+2)(r+3)}{6}{{x}^{r}}\]
C) \[\frac{(r+2)(r+3)}{2}{{x}^{r}}\]
D) None of these
Correct Answer: B
Solution :
\[{{(1-x)}^{-4}}\]\[=\left[ \frac{1.2.3}{6}{{x}^{0}}+\frac{2.3.4}{6}x+\frac{3.4.5}{6}{{x}^{2}}+\frac{4.5.6}{6}{{x}^{3}}+... \right.\]\[\left. +\frac{(r+1)(r+2)(r+3)}{6}{{x}^{r}}+.... \right]\] Therefore\[{{T}_{r+1}}=\frac{(r+1)(r+2)(r+3)}{6}{{x}^{r}}\].
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