Question

一道计算题求解

Answer 1

consider
(1+x)^2n =C(2n,0)x^(2n)+C(2n,1)x^(2n-1)+....+C(2n,2n)
x=1
2^(2n)=C(2n,0)+C(2n,1)+....+C(2n,2n) (1)
x=-1

0=C(2n,0)-C(2n,1)+....-C(2n,2n-1)+C(2n,2n) (2)
(1)-(2)
2[C(2n,2n-1)+C(2n,2n-3)+....+C(2n,2n-1) ] = 2^(2n)
C(2n,2n-1)+C(2n,2n-3)+....+C(2n,2n-1) = 2^(2n-1) (3)
2(1) - (3)
2C(2n,0) + C(2n,1)+2C(2n,2)+C(2n,3)+....+C(2n,2n-1) +2C(2n,2n)
=2^(2n+1) -2^(2n-1)
=3.2^(2n-1)