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已知数列Bn=1/(n^2+2n),其前n项和为Tn,求证Tn
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已知数列Bn=1/(n^2+2n),其前n项和为Tn,求证Tn
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答案和解析
Bn=1/(n^2+2n)=1/[n(n+2)]=1/2*[1/n-1/(n+2)]
Tn=1/2*(1-1/3)+1/2*(1/2-1/4)+1/2*(1/3-1/5)+1/2*(1/4-1/6)+...+1/2*[1/n-1/(n+2)]
=1/2*[1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+.+1/n-1/(n+2)]
=1/2[1+1/2-1/(n+1)-1/(n+2)]
Tn=1/2*(1-1/3)+1/2*(1/2-1/4)+1/2*(1/3-1/5)+1/2*(1/4-1/6)+...+1/2*[1/n-1/(n+2)]
=1/2*[1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+.+1/n-1/(n+2)]
=1/2[1+1/2-1/(n+1)-1/(n+2)]
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