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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).
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