函数f（x）=1/(4^x+m) (m>0),x1,x2属于R,当x1+x2=1时,f(x1)+f(x2)=1/2,求m的值

1.令x1=x2=1/2,f(1/2)=1/(2+m),f(1/2)+f(1/2)=1/2
因此1/(2+m)=1/4,m=2

2.x1+x2=1时,f(x1)+f(x2)=1/2，f(1/2)=1/4,an=f(0)+f(1/n)+f(2/n)+......+f(n/n)
因此当n为偶数2k时,k=n/2：
an=a(2k)
=[f(0/2k)+f(2k/2k)]+[f(1/2k)+f((2k-1)/2k)]+ … +[f((k-1)/2k)+f((k+1)/2k)]+f(k/2k)
=1/2+1/2+ … +1/2+f(1/2) ………… 有k个1/2
=k/2+1/4
=(n+1)/4

当n为奇数2k-1时,k=(n+1)/2：
an=a(2k-1)
=[f(0/(2k-1))+f((2k-1)/(2k-1))]+[f(1/(2k-1))+f((2k-2)/(2k-1)]+ … +[f((k-1)/(2k-1))+f(k/(2k-1))]
=1/2+1/2+ … +1/2+f(1/2) ………… 有k个1/2
=k/2
=(n+1)/4

综上，an=(n+1)/4