First slide

Binomial theorem for negative integral and rational Index

Question

If $(1 - x)^{- n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{r} x^{r} + \dots$ , then $a_{0} + a_{1} + a_{2} + \dots + a_{r}$ is equal to

Moderate

A

$\frac{n (n + 1) (n + 2) \dots (n + r)}{r!}$
B

$\frac{(n + 1) (n + 2) \dots (n + r)}{r!}$
C

$\frac{n (n + 1) (n + 2) \dots (n + r - 1)}{r!}$
D

$\frac{(n + 1) (n + 2) (n + 3) \dots (n + r)}{n!}$

Solution

We have,
$(1 - x)^{- n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{r} x^{r} + \dots$
and $(1 - x)^{- 1} = 1 + x + x^{2} + x^{3} + \dots + x^{r} + \dots$
Hence,
$\begin{matrix} a_{0} + a_{1} + a_{2} + \dots + a_{r} \\ = Coefficient of x^{r} in the product of the two series \\ = Coefficient of x^{r} in (1 - x)^{- n} (1 - x)^{- 1} \\ = Coefficient of x^{r} in (1 - x)^{- (n + 1)} \\ = \frac{(n + 1) (n + 2) \dots (n + r)}{r!} =^{n + r} C_{n} \end{matrix}$

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