已知数列{an}中a1=1 a[n+1]=3an 数列{bn}的前几项和Sn=n^2+2n,设cn=an*bn,求Tn=C1+C2=C3=-----Cn
问题描述：
已知数列{an}中a1=1 a[n+1]=3an 数列{bn}的前几项和Sn=n^2+2n,设cn=an*bn,求Tn=C1+C2=C3=-----Cn
答
a(n+1)/an=3,为定值
又a1=1,数列{an}是以1为首项,3为公比的等比数列.
an=1×3^(n-1)=3^(n-1)
n=1时,b1=S1=1+2=3
n≥2时,Sn=n²+2n S(n-1)=(n-1)²+2(n-1)
bn=Sn-S(n-1)=n²+2n-(n-1)²-2(n-1)=2n+1
n=1时,b1=2+1=3,同样满足.
数列{bn}的通项公式为bn=2n+1
cn=anbn=(2n+1)×3^(n-1)
Tn=c1+c2+...+cn=3×1+5×3+7×3²+...+(2n+1)×3^(n-1)
3Tn=3×3+5×3²+...+(2n-1)×3^(n-1)+(2n+1)×3ⁿ
Tn-3Tn=-2Tn=3+2×3+2×3²+...+2×3^(n-1) -(2n+1)×3ⁿ
=2×[1+3+3²+...+3^(n-1)]-(2n+1)×3ⁿ +1
=2×1×(3ⁿ-1)/(3-1) -(2n+1)×3ⁿ +1
=3ⁿ-1 -(2n+1)×3ⁿ +1
=-2n×3ⁿ
Tn=n×3ⁿ