在二项式(1+i)^（4n+2）的展开项中 各偶数项之和为多少?

(1+i)^(4n+2)二项式展开中的偶数项即多项式f(x) = (1+ix)^(4n+2)的偶次项系数.
f(x)的全体系数之和为:
f(1) = (1+i)^(4n+2) = ((1+i)^2)^(2n+1) = (2i)^(2n+1) = 2i·((2i)^2)^n = 2i·(-4)^n.
f(x)的偶次项系数之和减去奇次项系数之和为:
f(-1) = (1-i)^(4n+2) = ((1-i)^2)^(2n+1) = (-2i)^(2n+1) = -2i·((-2i)^2)^n = -2i·(-4)^n.
于是f(x)的偶次项系数之和为(f(1)+f(-1))/2 = 0.
也可以从复数乘方的角度看.
(1+i)^(4n+2)的偶数项系数和就是(1+i)^(4n+2)的实部.
设z = 1+i,其复共轭z' = 1-i = -iz.
因此z^(4n+2)的复共轭 = (z')^(4n+2) = (-iz)^(4n+2) = (-i)^(4n+2)·z^(4n+2) = -z^(4n+2).
由此可知(1+i)^(4n+2) = z^(4n+2)的实部为0.