Question

设f(x)=1/((2^x)+根号2),用推导等差数列前n项和公式的方法,求f(-5)+f(-4)+...+f(5)+f(6)的值求大神帮助

Answer 1

∵f(x) = 1/(2^x + √2） ∴f(1-x) = 1/[2^(1-x) + √2） ……(分子、分母同时乘以 2^x ) = 2^x/(2 + √2 * 2^x) ……（分母中提取出 √2） = (2^x/√2) ×(1/√2 + 2^x) = (2^x/√2) ×f(x) ∴f(x) + f(1-x) = (1+ 2^x/√2)× f(x) =[ (√2 + 2^x)/√2 ] ×f(x) = [1/√2f(x)] × f(x) = 1/√2 =√2/2 ∴f(-5)+f(-4)+……+f(0)+……+f(5)+f(6) = [f(-5) + f(6)] + [f(-4) + f(5)] + [f(-3) + f(4)] + [f(-2) + f(3)] + [f(-1) + f(2)] + [f(0) + f(1)] = 6×√2/2 =3√2