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已知1/n(n+1)=A/n+B/n+1求AB的值 再根据类比法计算 1/1*2+1/2*3+1/3*4+^+1/99*

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已知1/n(n+1)=A/n+B/n+1求AB的值 再根据类比法计算 1/1*2+1/2*3+1/3*4+^+1/99*100
1/n(n+1)=[(n+1)-n]/[n(n+1)]=(n+1)/[n(n+1)]-n/[n(n+1)]=1/n-1/(n+1)
又∵ 1/n(n+1)=A/n+B/(n+1)
∴ A=1,B=-1
∴ 1/[n(n+1)]=1/n-1/(n+1)
∴ 1/1*2+1/2*3+1/3*4+^+1/99*100
=1-1/2+1/2-1/3+1/3-1/4+.+1/99-1/100
=1-1/100
=99/100
再问: 1/a-1+1/(a-1)(a-2)+1/(a-2)(a-3)+^+1/(a-9)(a-10)? 急需
再答: 一样啊 1/a-1+1/(a-1)(a-2)+1/(a-2)(a-3)+^+1/(a-9)(a-10) =1/(a-1)+[1/(a-2)-1/(a-1)]+[1/(a-3)-1/(a-2)]+...............+[1/(a-10)-1/(a-9)] =1/(a-10)