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已知:正数数列{an}前n项和为Sn,满足6Sn=(an)^2+3an-10,求通项an (n是下标)
人气:209 ℃ 时间:2019-12-13 17:31:00
解答
n=1时6a1=(a1)^2+3a1-10 a1=5(负项舍去)6Sn=(an)^2+3an-106Sn-1=(an-1)^2+3an-1-10两式相减6an=(an)^2+3an-(an-1)^2-3an-1(an+an-1)(an-an-1)-3(an+an-1)=0an+an-1=0或者an-an-1-3=0由于是正项数列所以an+...
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