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

splitting up the series to two series, we get

a(C0-C1+C2...(-1)ⁿCn)+d(-C1+2C2...(-1)ⁿnCn)=a(1-x)ⁿ+d(d((1-x)ⁿ)/dx, where x is equal to unity,substituting, we get

 sum of each of the two series is zero. Therefore the answer is zero.

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Hi Himanshu,

Take a common and you have a{C0 - C1 + C2 -.......}     -------------- (1)

Take -b common and you have -b{C1 - 2C2 + 3C3 -........}  ------------- (2)

In the expansion of (1-x)^n, put x=1, you will get (1) = 0

And for the terms in bracket of (2), differentiate (1-x)^n, and put x=1

ie put x=1 in -n*(1-x)^n-1, which is again 0.

And hence the answer is 0.

Best Regards,

Ashwin (IIT Madras).