摘要:14.已知Sn为数列{an}的前n项和.且Sn与的等比中项为n(n∈N+).a1=.则Sn=_____.
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已知Sn为数列{an}的前n项和,且Sn=2an+n2-3n-2,n=1,2,3….
(Ⅰ)求证:数列{an-2n}为等比数列;
(Ⅱ)设bn=an•cosnπ,求数列{bn}的前n项和Pn;
(Ⅲ)设cn=
,数列{cn}的前n项和为Tn,求证:Tn<
.
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(Ⅰ)求证:数列{an-2n}为等比数列;
(Ⅱ)设bn=an•cosnπ,求数列{bn}的前n项和Pn;
(Ⅲ)设cn=
| 1 |
| an-n |
| 37 |
| 44 |
已知Sn为数列{an}的前n项和,且Sn=2an+n2-3n-2(n=1,2,3…).令bn=an-2n(n=1,2,3…).
(Ⅰ)求证:数列{bn}为等比数列;
(Ⅱ)令cn=
,记Tn=c1c2+2c2c3+22c3c4+…+2n-1cncn+1,比较Tn与
的大小.
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(Ⅰ)求证:数列{bn}为等比数列;
(Ⅱ)令cn=
| 1 |
| bn+1 |
| 1 |
| 6 |
已知Sn为数列{an}的前n项和,且Sn=2an+n2-3n-2(n∈N*)
(I)求证:数列{an-2n}为等比数列;
(II)设bn=an•cosnπ,求数列{bn}的前n项和Pn.
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(I)求证:数列{an-2n}为等比数列;
(II)设bn=an•cosnπ,求数列{bn}的前n项和Pn.