a1+3a2+9a3+……3^（n-1）an,求数列的通项公式

Question

a1+3a2+9a3+……3^（n-1）an,求数列的通项公式

Answer 1

a1+3a2+3^2a3+```````+3^(n-1)an=n/3
a1+3a2+3^2a3+```````+3^(n-2)a(n-1)=(n-1)/3
两式相减得
3^(n-1)an=n/3-(n-1)/3
3^(n-1)an=n/3-n/3+1/3
3^(n-1)an=1/3
3^nan=1
an=1/3^n
bn=nan=n/3^n
sn=1/3^1+2/3^2+3/3^3+.+n/3^n
sn/3=1/3^2+2/3^3+3/3^4+.+(n-1)/3^n+n/3^(n+1)
sn-sn/3=1/3^1+1/3^2+1/3^3+.+1/3^n-n/3^(n+1)
2sn/3=1/3*[1-(1/3)^n]/(1-1/3)-n/3^(n+1)
2sn/3=[1-(1/3)^n]/2-n/3^(n+1)
2sn/3=1/2-1/2*(1/3)^n-n/3^(n+1)
sn=3/4-3/4*(1/3)^n-3/2*n/3^(n+1)
sn=3/4-9/4*(1/3)^(n+1)-3/2*n/3^(n+1)
sn=3/4-(9/4+3/2*n)/3^(n+1)
sn=3/4-(9/4+3/2*n)/3^(n+1)
sn=3/4-3/4(3+2n)/3^(n+1)
sn=3/4-1/4(3+2n)/3^n