数列 (9 10:52:38)
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数列 (9 10:52:38)
社数列an是正数组成的数列,其前n项和为Sn,且对所有自然数n,an与2的等差中项等于Sn与2的等比中项,求数列{an}的通项公式
社数列an是正数组成的数列,其前n项和为Sn,且对所有自然数n,an与2的等差中项等于Sn与2的等比中项,求数列{an}的通项公式
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(an+2)/2=√2Sn
(an+2)^2=8Sn
an^2+4an+4=8Sn
an-1^2+4an-1+4=8Sn-1
两式相减
(an+an-1)(an-an-1)+4(an-an-1)=8an
[an+a(n-1)][an-a(n-1)]-4[an+a(n-1)]=0
[an+a(n-1)]{[an-a(n-1)]-4}=0
an为正数列;[an+a(n-1)]≠0
an-a(n-1)=4
在(an+2)^2=8Sn,令n=1,a1=2
所以an=4n-2
(an+2)^2=8Sn
an^2+4an+4=8Sn
an-1^2+4an-1+4=8Sn-1
两式相减
(an+an-1)(an-an-1)+4(an-an-1)=8an
[an+a(n-1)][an-a(n-1)]-4[an+a(n-1)]=0
[an+a(n-1)]{[an-a(n-1)]-4}=0
an为正数列;[an+a(n-1)]≠0
an-a(n-1)=4
在(an+2)^2=8Sn,令n=1,a1=2
所以an=4n-2