题目内容
已知递增等比数列{bn}满足b2•b4=64,b5=32,数列{an}满足an-bn=
.
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)设数列{cn}的通项公式cn=nan-
,求数列{cn}的前n项和Tn.
| 1 |
| 2n |
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)设数列{cn}的通项公式cn=nan-
| 1 |
| 2 |
(Ⅰ)∵递增等比数列{bn}满足b2•b4=64,b5=32,设公比为q,则有 b12 q5=64,且 b1q4=32,
解得 b1=2,q=2,bn=2n.
再由 {an}满足an-bn=
,得 an=bn+
=2n+
.
(Ⅱ)∵数列cn=nan-
,∴cn =n 2n.
∴数列{cn}的前n项和Tn=1×2+2×22+3×23+…+n•2n,①
2Tn=1×22+2×23+3×24+…+n•2n+1 ②.
①-②可得-Tn=2+22+23+…+2n-n•2n+1=2n+1-2-n•2n+1,
∴Tn=(n-1)2n+1+2.
解得 b1=2,q=2,bn=2n.
再由 {an}满足an-bn=
| 1 |
| 2n |
| 1 |
| 2n |
| 1 |
| 2n |
(Ⅱ)∵数列cn=nan-
| 1 |
| 2 |
∴数列{cn}的前n项和Tn=1×2+2×22+3×23+…+n•2n,①
2Tn=1×22+2×23+3×24+…+n•2n+1 ②.
①-②可得-Tn=2+22+23+…+2n-n•2n+1=2n+1-2-n•2n+1,
∴Tn=(n-1)2n+1+2.
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