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If $C_{r}$ stands for ${}^{n}C_{r},$ then the sum of first ( n + 1 ) terms of the $(n + 1)$ terms of the series $a C_{0} - (a + d) C_{1} + (a + 2 d) C_{2} - (a + 3 d) C_{3} + \dots$ , is

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Correct option is C

We have,aC0−(a+d)C1+(a+2d)C2−(a+3d)C3+…..........+(−1)n(a+nd)Cn =∑r−0n (a+rd)(−1)rnCr=a∑r=0n (−1)rnCr−dn∑r=1n−1 n−1Cr−1(−1)r1=a×0−dn×0=0

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If Cr stands for Cr n, then the sum of first ( n + 1 ) terms of the (n+1) terms of the series aC0−(a+d)C1+(a+2d)C2−(a+3d)C3+…, is

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If $C_{r}$ stands for ${}^{n}C_{r},$ then the sum of first ( n + 1 ) terms of the $(n + 1)$ terms of the series $a C_{0} - (a + d) C_{1} + (a + 2 d) C_{2} - (a + 3 d) C_{3} + \dots$ , is