计算:1/n(n+1)+1/(n+1)(n+2)+1/(n+2)(n+3)+1/(n+3)(n+4)
3个回答
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1/n(n+1)+1/(n+1)(n+2)+1/(n+2)(n+3)+1/(n+3)(n+4)
=1/n-1/(n+1)+1/(n+1)-1/(n+2)+1/(n+2)-1/(n+3)+1/(n+3)-1/(n+4)
=1/n-1/(n+4)
=(n+4-n)/[n(n+4)]
=4/[n(n+4)]
望采纳
=1/n-1/(n+1)+1/(n+1)-1/(n+2)+1/(n+2)-1/(n+3)+1/(n+3)-1/(n+4)
=1/n-1/(n+4)
=(n+4-n)/[n(n+4)]
=4/[n(n+4)]
望采纳
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展开全部
1/n(n+1)=1/n-1/(n+1)
因此
原式
=1/n-1/(n+1)+1/(n+1)-1/(n+2)+1/(n+2)-1/(n+3)+1/(n+3)-1/(n+4)
=1/n-1/(n+4)
=4/[n(n+4)]
因此
原式
=1/n-1/(n+1)+1/(n+1)-1/(n+2)+1/(n+2)-1/(n+3)+1/(n+3)-1/(n+4)
=1/n-1/(n+4)
=4/[n(n+4)]
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展开全部
1/n(n+1)+1/(n+1)(n+2)+1/(n+2)(n+3)+1/(n+3)(n+4)
=1/n-1/(n+1)+1/(n+1)-1/(n+2)+1/(n+2)-1/(n+3)+1/(n+3)-1/(n+4)
=1/n-1/(n+4)
=(n+4)/n(n+4)-n/n(n+4)
=(n+4-n)/n(n+4)
=4/n(n+4)
=1/n-1/(n+1)+1/(n+1)-1/(n+2)+1/(n+2)-1/(n+3)+1/(n+3)-1/(n+4)
=1/n-1/(n+4)
=(n+4)/n(n+4)-n/n(n+4)
=(n+4-n)/n(n+4)
=4/n(n+4)
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