f（x）=1/2^x+1 f(n)-1=1/2^n=(1/2)^n
f(-6)+f(-5)+……+f(0)+……+f(6)
=f(-6)+1+f(-5)+1+……+f(0)+……+f(6)+1-13
=2^6+2^5+……+2^1+1+(1/2)+……+(1/2)^6-13
=2^6[1-(1/2)^13]/(1-1/2)-13
=7359/64
=114又63/64额，，f（x）=1/(2^x+1)∵f(-6)=1/[2^(-6)+1]=2^6/(1+2^6) ∴f(-6)+f(6)=2^6/(1+2^6)+1/(2^6+1)=1同理：f(-5)+f(5) =f(-4)+f(4)=f(-3)+f(3)=f(-2)+f(2)=f(-1)+(1)=1∵f(0)=1/(2^0+1)=1/2∴f(-6)+f(-5)+……+f(0)+……+f(6)=6又2分之1.（即13/2)