设f(x)=1/[(2^x)+√2],则f(-5)+f(-4)+f(-3)+…+f(0)+f(1)+f(5)+f(6)=?正确答案是3√2球过程提问如题,

Question

设f(x)=1/[(2^x)+√2],则f(-5)+f(-4)+f(-3)+…+f(0)+f(1)+f(5)+f(6)=?正确答案是3√2球过程
提问如题,

Answer 1

因为f(x)+f(1-x)
=1/[(2^x)+√2]+1/[(2^1-x)+√2]
=1/[(2^x)+√2]+2^x/[2+√2(2^x)]
=√2/[√2(2^x)+2]+2^x/[2+√2(2^x)]
=(√2+2^x)/[2+√2(2^x)]=1/√2=√2/2,
所以f(-5)+f(6)=√2/2
f(-4)+f(5)=√2/2
f(-3)+f(4)=√2/2,
f(-2)+f(3)=√2/2,
f(-1)+f(2)=√2/2,
f(0)+f(1)=√2/2,
所以f(-5)+f(-4)+f(-3)+…+f(0)+f(1)+f(5)+f(6)=3√2.