f(x)=1/2^x+ 根号2, 利用求等差数列前n项和的公式的方法,求f(-5)+f(-4)……+f(0)+……+f(

f(x)=1/(2^x+√2),f(1-x)=1/[2^(1-x)+√2]=(2^x)/[ √2(√2+2^x)]
∴f(x)+ f(1-x)= 1/(2^x+√2)+ (2^x)/[ √2(√2+2^x)]=1/√2=(√2)/2
设S=f(-5)+f(-4)+…+f(5)+f(6)
则S= f(6)+f(5)+…+ f(-4)+f(-5)
2S=[ f(-5)+ f(6)]+[ f(-4)+ f(5)]+…+[ f(5)+ f(-4)]+[ f(6)+ f(-5)]
=12×(√2)/2=6√2
∴S=3√2,即f(-5)+f(-4)+…+f(5)+f(6)=3√2.