if a and d, are two complex numbers, then the sum to (n+ 1) terms of the following series aC0−(a+d)C1+(a+2d)C2−…+… is

We can write,aC0−(a+d)C1+(a+2d)C2−… upto (n+1) terms =aC0−C1+C2−…+d−C1+2C2−3C3+…We know,   (1−x)n=C0−C1x+C2x2−…+(−1)nCnxn…...(ii) On differentiating Eq. (ii) w.r.t. x, we get−n(1−x)n−1=−C1+2C2x−…+(−1)nCnnxn−1…...(iii) On putting x = 1 in Eqs. (ii) and (iii), we getC0−C1+C2−…+(−1)nCn=0-----(iv)and −C1+2C2−…+(−1)nnCn=0----(v)from eq.(i)aC0−(a+d)C1+(a+2d)C2−…upto (n+1) term=a⋅0+d⋅0=0 [fromEqs. (iv)and(v)]