This sum popped out of one of my calculations, I know what it should evaluate to, but I have no idea how to prove it.
r∑i=0(n2i)−(n2i−1)
I know that 2i−1 is negative for i=0, but for the purpose of this sum, we will say that (nx)=0 if n<0 or x<0. So this sum is basically summing the difference of consecutive even/odd binomial coefficient pairs. We can rewrite this sum as
2r∑i=0(−1)i(ni).
I don't really know how to proceed from here, I couldn't find any information on evaluating sums of alternating series involving binomial coefficients.
Answer
This is a special case of the result
m∑k=0(−1)k(nk)=(−1)m(n−1m)
(0≤m≤n−1) which can be proved by induction on m.
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