If $C_{r}$ denotes coefficient of $x^{r}$ in $(1 + x)^{99}$ , then the value of $C_{0} - 2 C_{1} + 3 C_{2} - 4 C_{3} + \dots - 100 C_{99}$ is

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Solution

$C_{0} - 2 C_{1} + 3 C_{2} - 4 C_{3} + \dots - 100 C_{99} = 99 \sum r = 0 (- 1)^{r} (r + 1) C_{r}$
(Here $C_{r} =^{99} C_{r}$ )
$= \sum r = 0 (- 1)^{r} r^{99} C_{r} + 99 \sum r = 0 (- 1)^{r} C_{r} = 99 \sum r = 1 (- 1)^{r} \cdot 99 \cdot^{98} C_{r - 1} + 99 \sum r = 0^{99} C_{r} (- 1)^{r} [∵ \frac{^{99} C_{r}}{^{98} C_{r - 1}} = \frac{99}{r}] = (- 1) 99 99 \sum r = 1^{98} C_{r - 1} (- 1)^{r - 1} + (1 - 1)^{99} = - 99 (1 - 1)^{98} + 0 = 0$

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