问题 解答题

已知数列{an}是等差数列,且a1=2,a1+a2+a3=12.

(1)求数列{an}的通项公式;

(2)令bn=an•3n,求数列{bn}的前n项和Sn

答案

(1)∵数列{an}是等差数列,且a1=2,a1+a2+a3=12,

∴2+2+d+2+2d=12,

解得d=2,

∴an=2+(n-1)×2=2n.

(2)∵an=2n,

∴bn=an•3n=2n•3n

∴Sn=2×3+4×32+6×33+…+2(n-1)×3n-1+2n×3n,①

3Sn=2×32+4×33+6×34+…+2(n-1)×3n+2n×3n+1,②

①-②得-2Sn=6+2×32+2×33+2×34+…+2×3n-2n×3n+1

=2×

3(1-3n)
1-3
-2n×3n+1

=3n+1-2n×3n+1

=(1-2n)×3n+1

∴Sn=

2n-1
2
×3n+1

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