an=2n-1,bn=2/3*[(1/3)^(n-1)]
∵cn=anbn
∴cn=2/3*(2n-1)(1/3)^(n-1)
∴Sn=2/3(1*1+3*1/3+5*(1/3)^2+……+(2n-3)*(1/3)^(n-2)+(2n-1)(1/3)^(n-1))
1/3Sn=2/3(1*1/3+3*(1/3)^2+5(1/3)^3+……+(2n-3)(1/3)^(n-1)+(2n-1)*(1/3)^n)
兩式相減得到
2/3Sn=2/3(1+2*1/3+2*(1/3)^2+……+2*(1/3)^(n-1)-(2n-1)*(1/3)^n)
∴Sn=2(1+1/3+(1/3)^2+……+(1/3)^(n-1))-(2n-1)(1/3)^n-1
=3(1-(1/3)^n)-(2n-1)(1/3)^n-1
=2-(2n-1+3)(1/3)^n
=2-2(n+1)(1/3)^n