排列组合题：C(2n,2)+C(2n,4)+……+C(2n,2k)+……+C(2n,2k)=？

由二项式定理 (1+1)^n=C(2n,0)+C(2n,1)+……+C(2n,2k)+……+C(2n,2n) (1-1)^n=C(2n,0)-C(2n,1)+……-C(2n,2k-1)+C(2n,2k)+……+C(2n,2n) 相加 2^n =C(2n,0)+ C(2n,2)+C(2n,4)+……+C(2n,2k)+……+C(2n,2n) C(2n,2)+C(2n,4)+……+C(2n,2k)+……+C(2n,2n)=2^n-C(2n,0) = 2^n-1