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设数列{an}的前n项和Sn,且a1=1,Sn=4an+2(n∈N*)
1.设bn=an+1-2an,求证:{bn}是等比数列
2.设Cn=an/2∧n,求证:{Cn}是等差数列
3.求Sn=a1+a2+...+an
人气:173 ℃ 时间:2020-06-03 15:19:25
解答
题目中应该是:S(n+1)=4an+2∵S(n+1)=4an+2Sn=4a(n-1)+2∴S(n+1)-Sn=4an-4a(n-1)a(n+1)=4an-4a(n-1)a(n+1)-2n=2(an-2a(n-1))a(n+1)-2n/(an-2a(n-1))=2∵bn/b(n-1)=a(n+1)-2n/(an-2a...
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