已知数列an满足a1,a2-a1,a3-a2,…,an-a(n-1)是首项为1公比为1/3的等比数列,①求an的通项公式②若bn=(2n-1)an,求数列bn的前n项和sn

Question

已知数列an满足a1,a2-a1,a3-a2,…,an-a(n-1)是首项为1公比为1/3的等比数列,①求an的通项公式②若bn=(2n-1)an,求数列bn的前n项和sn

Answer 1

设数列a1,a2-a1,a3-a2.an-a(n-1) 和为Tn
Tn=a1+a2-a1+a3-a2.an-a(n-1)=an=3/2-3(1/3)^n/2
bn=（2/3)n[3/2-3(1/3)^n/2]
=n-n(1/3)^n
设Sn=S1-S2
S1=1+2+3+……+n=n(1+n)/2
S2=1/3+2*(1/3)^2+3*(1/3)^3+4*(1/3)^4+……+n(1/3)^n
S2/3=(1/3)^2+2*(1/3)^3+3*(1/3)^4+4*(1/3)^5+……+n(1/3)^(n+1)
S2-S2/3=1/3+(1/3)^2+(1/3)^3+(1/3)^4+……+(1/3)^n-n(1/3)^(n+1)
2S2/3=1/2-(1/3)^n/2-n(1/3)^(n+1)
S2=3/4-3(1/3)^n/4-3n(1/3)^(n+1)/2
Sn=n(1+n)/2-3/4+3(1/3)^n/4+3n(1/3)^(n+1)/2