If $a$ and $d$ are two complex numbers, then the sum to $(n + 1)$ terms of the following series $a C_{0} - (a + d) C_{1} + (a + 2 d) C_{2} - . . . . + . . . . .$ is

\frac{a}{2 n}

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n a

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0

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none of these

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Solution

The correct option is C $0$
Given expression is $a C_{0} - (a + d) C_{1} + (a + 2 d) C_{2} - . . . . + . . . . .$
Simplifying the above expression, we get
$a (^{n} C_{0} -^{n} C_{1} +^{n} C_{2} -^{n} C_{3} + . . . (- 1)^{n}^{n} C_{n}) - d (^{n} C_{1} - 2^{n} C_{2} + 3^{n} C_{3} - 4^{n} C_{4} + . . . n (- 1)^{n}^{n} C_{n})$
$= a (1 + x)_{x = - 1}^{n} - d {(\frac{d (1 + x)^{n}}{d x})}_{x = - 1}$
$= 0$

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