Q. The $(r + 1)$th term in the expansion of $(1 - x)^{- 4}$ will be
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
Explanation
(b)
$(1 - x)^{- 4} = 1.x^{0} + {4x}^{1} + \frac{4.5}{2}x^{2} + ...$
$= \left\lbrack \frac{1.2.3}{6}x^{0} + \frac{2.3.4}{6}x + \frac{3.4.5}{6}x^{2} + \frac{4.5.6}{6}x^{3} + ... + \frac{(r + 1)(r + 2)(r + 3)}{6}x^{r} + ... \right\rbrack$
Therefore, $T_{r + 1} = \frac{(r + 1)(r + 2)(r + 3)}{6}x^{r}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.