已知函数f(x)=1⼀4^x+2,计算f(0)+f(1⼀10)+f(2⼀10)+…+f(9⼀10)+f(1)

题目没有问题
f(x)=1/(4^x+2)
f(x)+f(1-x)
=1/(4^x+2)+1/[4^(1-x)+2]
=1/(4^x+2)+1/[4/4^x+2]
=1/(4^x+2)+4^x/(4+2*4^x)
=1/(4^x+2)+4^x/[2(4^x+2)]
=(2+4^x)/[2(4^x+2)]
=1/2
∴f(0)+f(1)=1/2
f(1/10)+f(9/10)=1/2
.............................
f(4/10)+f(6/10)=1/2
f(5/10)=1/4
相加：
f(0)+f(1/10)+f(2/10)+…+f(9/10)+f(1)
=5×1/2+1/4
=11/4参考http://58.130.5.100/

题目有问题？应该是函数f(x)=4^x/(4^x+2)
则f(x)+f(1-x)=4^x/(4^x+2)+4^(1-x)/[4^(1-x)+2]
=4^x/(4^x+1)+4/(4+2*4^x]
=4^x/(4^x+2)+2/(4^x+2)
=(4^x+2)/(4^x+2)
=1 (注意：f(x)+f(1-x)=1 其中x+1-x=1)
所以f(0)+f(1/10)+f(2/10)+…+f(9/10)+f(1)
=[f(0)+f(1）]+[f(1/10)+f(9/10)+...+[f(4/10)+f(6/10)]+f(5/10)
=1+1+1+1+1+(1/2)
=5.5
希望能帮到你O(∩_∩)O

f(0)+f(1/10)+f(2/10)+…+f(9/10)+f(1)
=[1/4^0+1/4^(0.1)+1/4^0.2+1/4^0.3+...+1/4^0.9+1/4^1]+2+2+...+2
=[1(1-(1/4^0.1)^10]/(1-1/4^0.1) +2*10
=(1-1/4)/(1-1/4^(1/10))+20
=3/4/(1-4^(-0.1))+20
=3/[4-4^(0.9)]+20