设{an}公差为d,设{bn}公比为q
S5-2a1=5a1+10d-2a1
=3a1+10d
=3a1+6d+4d
=3(a1+2d)+4d
=3a3+4d
=3×3+4d=17
d=(17-3×3)/4=2
an=a1+(n-1)d=a3+(n-3)d=3+2(n-3)=2n-3
b1=a2=2×2-3=1
b2S3=6 b2=6/S3=6/(3a2)=2/a2=2/1=2
q=b2/b1=2/1=2
bn=b1q^(n-1)=1×2^(n-1)=2^(n-1)
cn=a(n+1)bn=[2(n+1)-3]×2^(n-1)=(2n-1)×2^(n-1)
Tn=c1+c2+...+cn=1×1+3×2+5×2^2+...+(2n-1)×2^(n-1)
2Tn=1×2+3×2^2+...+(2n-3)×2^(n-1)+(2n-1)×2^n
Tn-2Tn=-Tn=1+2×2+2×2^2+...+2×2^(n-1)-(2n-1)×2^n
=2[1+2+...+2^(n-1)] -(2n-1)×2^n -1
=2×1×(2^n -1)/(2-1) -(2n-1)×2^n -1
=(3-2n)×2^n -3
Tn=(2n-3)×2^n +3