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×
-2n×3n+13(1-3n) 1-3
=3n+1-2n×3n+1
=(1-2n)×3n+1
∴Sn=
×3n+1.2n-1 2