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求和:Sn=1/(1^2+3)+1/(2^2+6)+1/(3^2+9)+.+1/(n^2+3n)

分类: 作业答案 • 2022-01-07 11:55:44
问题描述：

求和:Sn=1/(1^2+3)+1/(2^2+6)+1/(3^2+9)+.+1/(n^2+3n)

答

1/(n^2+3n)=1/n(n+3)=[1/n-1/(n+3)]/3
Sn=1/(1^2+3)+1/(2^2+6)+1/(3^2+9)+.+1/(n^2+3n)
=[1-1/4+1/2-1/5+1/3-1/6+1/4-1/7+-----+1/(n-1)-1/(n+2)+1/n-1/(n+3)]/3
=[1+1/2+1/3-1/(n+1)-1/(n+2)-1/n+3)]
=11/6-(3n^2+12n+11)/(n^3+6n^2+11n+6)

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