问题
解答题
数列{an}的前n项和为Sn,且a1=2,an+1=
(1)若等差数列{bn}恰好使数列{an+bn}成公比为
(2)求通项an (3)求
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答案
(1)因为a1=2,an+1=
1 |
3 |
5 |
3 |
所以an+1-3(n+1)+2=
1 |
3 |
所以数列{an-3n+2}以1为首项,
1 |
3 |
所以bn=-3n+2时,等差数列{bn}恰好使数列{an+bn}成公比为
1 |
3 |
(2)由(1)可知数列{an-3n+2}以1为首项,
1 |
3 |
所以an-3n+2=(
1 |
3 |
1 |
3 |
(3)由(2)可知,数列{an}的前n项和为:
Sn=
1-(
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1-
|
3n(1+n) |
2 |
3 |
2 |
3 |
2 |
1 |
3 |
3n(1+n) |
2 |
∴
lim |
n→∞ |
Sn |
n2 |
lim |
n→∞ |
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n2 |
3 |
2 |