an |
3n |
an+1 |
3n+1 |
∴bn+1−bn=
an+1 |
3n+1 |
an |
3n |
1 |
3 |
所以数列{bn}是等差数列,首项b1=1,公差为
1 |
3 |
∴bn=1+
1 |
3 |
n+2 |
3 |
(2)an=3nbn=(n+2)×3n−1-------------------------(7分)
∴Sn=a1+a2+…+an=3×1+4×3+…+(n+2)×3n-1----①
∴3Sn=3×3+4×32+…+(n+2)×3n-------------------②(9分)
①-②得−2Sn=3×1+3+32+…+3n−1−(n+2)×3n
=2+1+3+32+…+3n-1-(n+2)×3n=
3n+3 |
2 |
∴Sn=−
3n+3 |
4 |
(n+2)3n |
2 |