由③知f(0)=2f(0)-2⇒f(0)=2;
(Ⅱ)任取x1x2∈[0,1],且x1<x2,
则0<x2-x1≤1,∴f(x2-x1)≥2
∴f(x2)-f(x1)=f[(x2-x1)+x1]-f(x1)
=f(x2-x1)+f(x1)-2-f(x1)=f(x2-x1)-2≥0
∴f(x2)≥f(x1),则f(x)≤f(1)=3.
∴f(x)的最大值为3;
(Ⅲ)由Sn=−
1 |
2 |
当n=1时,a1=1;当n≥2时,an=−
1 |
2 |
1 |
2 |
∴an=
1 |
3 |
1 |
3n−1 |
∴f(an)=f(
1 |
3n−1 |
1 |
3n |
1 |
3n |
1 |
3n |
2 |
3n |
1 |
3n |
=3f(
1 |
3n |
∴f(an+1)=
1 |
3 |
4 |
3 |
∴f(an+1)−2=
1 |
3 |
又f(a1)-2=1∴f(an)−2=(
1 |
3 |
1 |
3 |
∴f(a1)+f(a2)++f(an)=2n+
3 |
2 |
1 |
2×3n−1 |