1.设数列首项为a1,公差为d
a1+a2+..an=n^2
-a1+a2+...-a(n-1)+an=n
d*n/2=n
d=2
na1+n(n-1)d/2=n^2
n^2-n+na1=n^2
a1=1
an=1+2(n-1)=2n-1
2.f(1/2)=1*1/2+3*(1/2)^2+...(2n-1)*(1/2)^n
1/2f(1/2)= 1*(1/2)^2+...(2n-3)*(1/2)^n+(2n-1)*(1/2)^(n+1)
1/2f(1/2)=1/2+2[(1/2)^2+..(1/2)^n]-(2n-1)*(1/2)^(n+1)
=1/2+[1-(1/2)^(n-1)]-(2n-1)*(1/2)^(n+1)
f(1/2)=3-(1/2)^(n-2)-(2n-1)*(1/2)^n