f(x)=1/(2^z+1),求f(-5)+f(-4)+f(-3)+……+f(4)+f(5)的值

f(x)=1/(2^z+1),求f(-5)+f(-4)+f(-3)+……+f(4)+f(5)的值
f(x)=1/(2^z+1),求f(-5)+f(-4)+f(-3)+……+f(4)+f(5)的值

f(x)=1/(2^z+1),求f(-5)+f(-4)+f(-3)+……+f(4)+f(5)的值
f(-z)=1/[2^(-z)+1]=2^z/(1+2^z)
所以f(z)+f(-z)=1/(2^z+1)+2^z/(1+2^z)=1
所以f(-5)+f(5)=f(-4)+f(4)=…=f(-1)+f(1)=1
而f(0)=1/(1+1)=1/2
所以原式=1+1+1+1+1+1/2=11/2