函数fx=1/4^x+m(m>0),当x1+x2=1时,fx1+fx2=1/2,
问题描述：
函数fx=1/4^x+m(m>0),当x1+x2=1时,fx1+fx2=1/2,
1,求m的值
2,已知数列an满足an=f0+f(1/n)+f(2/n)+……+f(n-1/n),求an
3,若sn=a1+a2+a3+……+an,求sn
4^x+m是分母 x1,x2属于R
答
1.fx1=1/(4^x1+m)fx2=1/(4^x2+m)fx1+fx2=[(4^x1+4^x2)+2m]/[4^(x1+x2)+m(4^x1+4^x2)+m^2]令t=4^x1+4^x2,化简得fx1+fx2=(t+2m)/(4+mt+m^2)=1/22t+4m=4+mt+m^2m^2+(t-4)m+4-2t=0(m-2)(m-2+t)=0m=2 或 m=2-t=2-4^x1-4^x...