求数列1/2^2+4,1/4^2+8,1/6^2+12.1/(2n)^2+4n的前n项和和sn
人气：299 ℃ 时间：2020-03-28 11:36:13
解答
1/{2^2+4]+ 1[/4^2+8]+1/[6^2+12]+...+1/[(2n)^2+4n]consider1/[(2n)^2+4n]=1/[4n(n+1)]=(1/4)[1/n -1/(n+1)]1/{2^2+4]+ 1[/4^2+8]+1/[6^2+12]+...+1/[(2n)^2+4n]=(1/4) [ (1/1-1/2)+(1/2-1/3)+...+(1/n-1/(n+1)]=(1...
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