设数列{bn}的前n项和为Sn,且bn=2-2s.数列{an}为等差数列,且a5=14,a7=20.

∵bn=2-2Sn,∴b[n-1]=2-S[n-1]
则bn-b[n-1]=-2(Sn-S[n-1])=-2bn
∴3bn=b[n-1]
即bn/b[n-1]=1/3,
b1=2-2b1,得b1=2/3
{bn}是b1=2/3,q=1/3的等比数列
∴bn=2/3*(1/3)^(n-1)=2/3^n,
∵a5=14,a7=20
a7=a5+2d=20
∴d=3
∵a5=a1+4d
∴a1=2
又∵（an)是等差数列
∴an=2+3(n-1)
=3n-1
cn=(3n-1)*2/3^n
Tn=2*2/3+5*2/3^2+8*2/3^3+.+(3n-1)*2/3^n
3Tn=2*2+5*2/3+8*2/3^2+.+(3n-1)*2/3^(n-1)
错位相减得
Tn=7/2-3/2•(1/3)^(n-1)- (3n-1)(1/3)^n