设f(x)= 1/(2^ x+根号2),利用课本中推导等差数列前n项和的公式

f(x) = 1/[2^x + 2^(1/2)],
f(1-x) = 1/[2^(1-x) + 2^(1/2)] = 2^x/[2 + 2^(1/2 + x)] 分子分母同乘2^x
= 2^(x-1/2)[2^x + 2^(1/2)] 分子分母同除2^(1/2)
f(x) + f(1-x) = [1 + 2^(x-1/2)]/[2^x + 2^(1/2)]
= 2^(-1/2)[ 2^x + 2^(1/2)]/[2^x + 2^(1/2)] 分子提出2^(-1/2)因子
= 2^(-1/2)
= 1/2^(1/2)
不等于1哈,但思路应该是一样的.