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数列an的前n项和为Sn.且满足a1=1.2Sn=(n+1)an
(1)求an与a(n-1)的关系式.并求an的通项公式
(2)求和Wn=1/(a2^2-1)+1/(a3^2-1)+……+1/(a(n+1)^2-1)
人气:358 ℃ 时间:2019-08-31 09:06:22
解答
2·a(n) = 2[Sn - S(n-1)] = (n+1)an - n·a(n-1)∴(n-1)an = n·a(n-1) ,∴an/[a(n-1)] = n/(n-1) ,.,a3/a2 = 3/2 ,a2/a1 = 2/1 ,将上述式子相乘:an/a1 = n ,∴an = n·1 = n ,即:{an}是以1为首项、1为公差的等差...
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