a−nC1(a−1)+nC2(a−2)− …+(−1)n(a−n) =a(nC0−nC1+nC2−nC3+…+(−1)nnCn) +(nC1−2nC2+3nC3−…+(−1)n+1nnCn) ...(i)
As we know (1−x)n=nC0−xnC1+x2nC2−x3nC3+ …+(−1)nxnnCn ...(ii)
Put x=1, we get 0=nC0−nC1+nC2−nC3+…+(−1)nnCn
On differentiating Eq. (ii) w.r.t. x, we get n(1−x)n−1=−nC1+2xnC2−3x2nC3+… +(−1)nnxn−1nCn
Put x=1 0=−nC1+2nC2−3nC3+….+(−1)n−1nnCn
From Eq. (i) a−nC1(a−1)+nC2(a−2)−…+(−1)n(a−n) =a(0)+0=0