已知an,bn的通项公式,怎么求Cn=an*bn,｛Cn｝的前n项和例如,an=-2/3*(1/3)^n-1,bn=2n-1

Question

已知an,bn的通项公式,怎么求Cn=an*bn,｛Cn｝的前n项和
例如,an=-2/3*(1/3)^n-1,bn=2n-1

Answer 1

｛Cn｝:Cn=an*bn=[-2/3*(1/3)^(n-1)]*(2n-1）
前n项和为Sn=-2/3*[1*(1/3)^0+3*(1/3)^1+...+(2n-1)*(1/3)^(n-1)]
1/3Sn=-2/3*[1*(1/3)^1+3*(1/3)^2+...+(2n-1)*(1/3)^n]
2/3Sn=Sn-1/3Sn=-2/3*[1*(1/3)^0+2*(1/3)^1+...+2*(1/3)^(n-1)-(2n-1)*(1/3)^n]
=-2/3*{1-(2n-1)*(1/3)^n+2*[(1/3)^1+(1/3)^2+...+(1/3)^(n-1)]}
Sn=-{1-(2n-1)*(1/3)^n+2*(1/3)*[1-(1/3)^(n-1)]/(1-1/3)}
=2-2*(n+1)*(1/3)^n