Find the sum of the series
∑r=0n(−1)r nCr[12r+3r22r+7r23r+15r24r⋯upto m terms]
2mn−12mn(2n−1)
=∑nr=0(−1)r nCr(12)r+∑nr=0(−r)r nCr(34)r+∑nr=0(−1)r nCr(78)r+⋯ upto m terms
=(1−12)n+(1−34)n+(1−78)n+⋯ upto m terms
[using ∑nr=0(−1)r nCr.xr=(1−x)n]
=(12)n+(14)n+(18)n+⋯ upto m terms
=(12)n[1−(12n)m1−12n]=2mn−12mn(2n−1)