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

\frac{n (n + 1) (n + 2) . . . . (n + r)}{r!}

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\frac{(n + 1) (n + 2) . . . . (n + r)}{r!}

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\frac{n (n + 1) (n + 2) . . . . (n + r - 1)}{r!}

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None of the above

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Solution

The correct option is B $\frac{(n + 1) (n + 2) . . . . (n + r)}{r!}$
we have,
$(1 - x)^{- n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . . . + a_{r} x^{r} + . . . . .$
and
$(1 - x)^{- 1} = 1 + x + x^{2} + x^{3} + . . . . + x^{r} + . . . . .$
Hence,
$a_{0} + a_{1} + a_{2} + . . . . . . + a_{r}$
$=$ coefficient of $x^{r}$ in the product of two series
$=$ coefficient of $x^{r}$ in $(1 - x)^{- n} (1 - x)^{- 1}$
$=$ coefficient of $x^{r}$ in $(1 - x)^{- (n + 1)}$
$= \frac{(n + 1) (n + 2) . . . . (n + r)}{r!}$

Alternate Solution:
General term in the expansion of $(1 - x)^{- n}$ will be
$T_{r + 1} =^{n + r - 1} C_{r} x^{r}$
hence $a_{0} + a_{1} + a_{2} + . . . . . . + a_{r}$
$=^{n - 1} C_{0} +^{n} C_{1} +^{n + 1} C_{2} + \dots +^{n + r - 1} C_{r}$
Now as we know that
$^{n - 1} C_{0} +^{n} C_{1} =^{n} C_{0} +^{n} C_{1} =^{n + 1} C_{1}$
similarly $^{n + 1} C_{1} +^{n + 1} C_{2} =^{n + 2} C_{2}$
Hence
$^{n - 1} C_{0} +^{n} C_{1} +^{n + 1} C_{2} + \dots +^{n + r - 1} C_{r} =^{n + r} C_{r} = \frac{(n + r)!}{n! \cdot r!} = \frac{(n + r) (n + r - 1) \dots (n + 2) (n + 1)}{r!}$

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