摘要:设数列{an}的前n项和为Sn且a1=1.Sn=nan-2n(n-1)(nÎN*).(1) 求证:数列{an}为等差数列.并求Sn,
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设数列{an}的前n项和为Sn,且a1=1,Sn+1=4an+2(n∈N*).
(1)设bn=an+1-2an,求证:数列{bn}是等比数列;
(2)设cn=
,求证:数列{cn}是等差数列.
设数列{an}的前n项和为Sn且Sn2-2Sn-anSn+1=0,n=1,2,3…
(1)求a1,a2
(2)求Sn与Sn-1(n≥2)的关系式,并证明数列{
}是等差数列.
(3)求S1•S2•S3…S2010•S2011的值.
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(1)求a1,a2
(2)求Sn与Sn-1(n≥2)的关系式,并证明数列{
| 1 | Sn-1 |
(3)求S1•S2•S3…S2010•S2011的值.