Question

设数列{an}满足a1+3a2+3^2a3+```````+3^(n-1)an=n/3(n属于正整数） (1)求数列{an}的通项公式

（2）bn=nan,求数列{bn}的前n项和sn

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

Answer 2

1
an= --------
3^n

n(n+1)
sn= ----------------
3(3^n -1)