江湖救急...
已知数列{an}各项均为正数其前n项和为Sn且满足4Sn=(an+1)平方
求{an}的通项公式.
人气:424 ℃ 时间:2020-06-30 06:15:25
解答
4sn=(an+1)^2
4S(n-1)=(a(n-1)+1)^2
4Sn-4S(n-1)=4an=an^2+2an+1-(a(n-1)+1)^2
(an-1)^2=(a(n-1)+1)^2
an-1=a(n-1)+1 或an-1=-a(n-1)-1舍去
an-a(n-1)=2
4s1=(a1+1)^2=a1
a1=1
an=2n-1
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