等差数列an前n项和为Sn,已知对任意n∈N+,点(n,Sn)在二次函数f(x)=x^2+c的图像上,（1）求c,an,（2）若kn=an/(2^n),求数列kn前n项和Tn.
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sn=n^2+c
s1=a1=1^2+c
a1=1+c
sn=n^2+c
s(n-1)=(n-1)^2+c
两式相减得an=2n-1
a1=2*1-1=1
1+c=1c=0kn=an/2^n
=2n/2^n
=n/2^(n-1)
Stn=1/2^0+2/2^1+.+n/2^(n-1)
Stn/2=1/2^1+2/2^2+.+n/2^n
Stn-Stn/2=1/2^0+1/2^1+1/2^2+.+1/2^(n-1)-n/2^n
Stn/2=[1-(1/2)^n]/(1-1/2)-n/2^n
Stn/2=2*[1-(1/2)^n]-n/2^n
Stn/2=2-2/2^n-n/2^n
Stn/2=2-(n+2)/2^n
Stn=4-(n+2)/2^(n-1)