题目内容
等差数列{an}中,a2+a3+a4=15,a5=9.
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)设bn=3
,求数列{
×bn}的前n项和Sn.
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)设bn=3
| an+1 |
| 2 |
| an+1 |
| 2 |
(Ⅰ)设数列{an}的公差为d首项为a1,由题意得,
,即
,
解得a1=1,d=2,
∴数列{an}的通项公式an=2n-1,
(Ⅱ)由(Ⅰ)可得bn=3
=3n,∴
×bn=n3n,
∴Sn=1×3+2×32+3×33+…+n×3n,①
3Sn=1×32+2×33+3×34+…+n×3n+1,②
①-②得,-2Sn=3+32+33+…+3n-n×3n+1=
-n×3n+1
=
-n×3n+1,
∴Sn=
.
|
|
解得a1=1,d=2,
∴数列{an}的通项公式an=2n-1,
(Ⅱ)由(Ⅰ)可得bn=3
| an+1 |
| 2 |
| an+1 |
| 2 |
∴Sn=1×3+2×32+3×33+…+n×3n,①
3Sn=1×32+2×33+3×34+…+n×3n+1,②
①-②得,-2Sn=3+32+33+…+3n-n×3n+1=
| 3(1-3n) |
| 1-3 |
=
| 3(3n-1) |
| 2 |
∴Sn=
| 3+(2n-1)•3n+1 |
| 4 |
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