f(1)=2m+t=1 f(2)=4m+t=3 m=1 t=-1 f(x)=2^x-1 Sn=2^n-1
a1=S1=1.当n>=2时,an=Sn-S(n-1)=2^n-2^(n-1)=2^(n-1),a1=1也适合.
所以an=2^(n-1)(n=1,2,3,……,).cn=6nan-n=6n*2^(n-1)-n
Tn=6(1*2^0+2*2^1+3*2^2+…+n*2^(n-1))-(1+2+3+…+n)
设An=1*2^0+2*2^1+3*2^2+…+n*2^(n-1) (1) Bn=1+2+3+…+n=n(n+1)/2
2*(1)得：2An=1*2^1+2*2^2+3*2^3+…+n*2^n (2)
(1)-(2)得：-An=1+2+2^2+2^3+…+2^(n-1)-n*2^n=2^n-1-n*2^n
An=(n-1)*2^n+1
Tn=6An-Bn=6(n-1)*2^n+1-n(n+1)/2