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求和:Sn=2^/1*3+4^2/3*5+……(2n)^2/(2n-1)(2n+1)
人气:245 ℃ 时间:2020-03-26 04:35:55
解答
Sn=2^/1*3+4^2/3*5+……(2n)^2/(2n-1)(2n+1)=1+1/1*3+1+1/3*5+...+1+1/(2n-1)(2n+1)=n+(1/1*3+1/3*5+...+1/(2n-1)(2n+1))=n+[(1-1/3)/2+(1/3-1/5)/2+...+(1/(2n-1)-1/(2n+1)/2]=n+(1-1/(2n+1))/2=n+(2n/(2n+1)/2=n+n...
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