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Sum of Product of Binomial Coefficients

If n is an in...

Question

If n is an integer greater than 1, then $a -^{n} C_{1} (a - 1) +^{n} C_{2} (a - 2) - . . . . + (- 1)^{n} (a - n) =$

a

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0

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a^{2}

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2^{n}

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Solution

The correct option is B $0$
$a -^{n} C_{1} (a - 1) +^{n} C_{2} (a - 2) - . . . \dots . + (- 1)^{n} (a - n)$
$= n \sum r = 0 (- 1)^{r}^{n} C_{r} (a - r)$
$= n \sum r = 0 (- 1)^{r}^{n} C_{r} \cdot a - n \sum r = 0 (- 1)^{r}^{n} C_{r} \cdot r$
$= a n \sum r = 0 (- 1)^{r}^{n} C_{r} - n \sum r = 1 (- 1)^{r}^{n} C_{r} \cdot r$ (when $r = 0$ , we get $0$ )
$= a (1 - 1)^{n} - n \sum r = 1 (- 1)^{r} \cdot \frac{n}{r}^{n - 1} C_{r - 1} \cdot r$
$(^{n} C_{r} = \frac{n}{r}^{n - 1} C_{r - 1})$ $((1 + x)^{n} = n \sum r = 0^{n} C_{r} x^{r}$ when $x = - 1$ $(1 - 1)^{r} = n \sum r = 0^{n} C_{r} (- 1)^{r})$
$= a (0) - n \sum r = 1 (- 1)^{r} n \cdot^{n - 1} C_{r - 1}$
$= 0 - n n \sum r = 1 (- 1)^{r}^{n - 1} C_{r - 1}$
$= - n [-^{n - 1} C_{0} +^{n - 1} C_{1} -^{n - 1} C_{2} \dots \dots \dots (- 1)^{n - 1}^{n - 1} C_{n - 1}]$
$= - n [0] = 0$ $[(1 - 1)^{n - 1} =^{n - 1} C_{0} -^{n - 1} C_{1} +^{n + 1} C_{2} -^{n - 1} C_{3} + . . . . . = (- 1)^{n}^{n - 1} C_{n - 1} = 0]$ .

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