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幼苗
共回答了23个问题采纳率:87% 举报
用裂项法
1/[n(n+k)]=1/k[1/n-1/(n+k)].
数列{1/[n(n+k)]}前n项和为:
1/[1(1+k)]+ 1/[2(2+k)]+ 1/[3(3+k)]+……+1/[n(n+k)]
=1/k[1-1/(1+k)+1/2- 1/(2+k)+1/3-1/(3+k)+……+1/n-1/(n+k)]
=1/k[(1+1/2+1/3+……+1/n)-( 1/(1+k)+ 1/(2+k)+ 1/(3+k)+……+1/(n+k)]
=1/k[(1+1/2+1/3+……+1/k)- ( 1/(n+1)+ 1/(n+2)+ 1/(n+3)+……+1/(n+k)]
比如k=2时,数列{1/[n(n+2)]}前n项和为:
1/2[(1+1/2)-(1/(n+1)+1/(n+2) ].
1年前
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