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

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Solution

$x^{r}$ coefficient in $(1 + x)^{99} =^{99} C_{r}$
So $C_{r}$ represents here $^{99} C_{r}$

$(1 + x)^{99} = C_{0} + C_{1} x + C_{2} x^{2} + C_{3} x^{3} + \dots + C_{99} x^{99}$
$\Rightarrow x (1 + x)^{99} = C_{0} x + C_{1} x^{2} + C_{2} x^{3} + C_{3} x^{4} + \dots + C_{99} x^{100} =$
Differentiating wrt x on both sides

$99 x (1 + x)^{98} + (1 + x)^{99} = C_{0} + 2 C_{1} x + 3 C_{2} x^{2} + 4 C_{3} x^{3} + \dots + (100) C_{99} x^{99} =$
By putting $x = - 1,$ we get
$C_{0} - 2 C_{1} + 3 C_{2} - 4 C_{3} + \dots 100 C_{99} = 0$

Alternate 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 $C_{r} =^{99} C_{r}$ )
$= \sum r = 0 (- 1)^{r} r^{99} C_{r} + 99 \sum r = 0 (- 1)^{r} C_{r} = 99 \sum r = 1 (- 1)^{r} \cdot 98 \cdot^{98} C_{r - 1} + 99 \sum r = 0^{99} C_{r} (- 1)^{r}$ $[∵ \frac{^{99} C_{r}}{^{98} C_{r - 1}} = \frac{99}{r}]$
$= (- 1) 98 99 \sum r = 1^{98} C_{r - 1} (- 1)^{r - 1} + 0$ $[∵ n \sum r = 0 (a)^{r}^{n} C_{r} = (1 + a)^{n}]$
$= 0$

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