(1/1*2)+(1/2*3)+(1/3*4)……+(1/99*100)
来源:百度知道 编辑:UC知道 时间:2024/06/17 13:35:40
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能不能写得更详细点,谢谢!
能不能写得更详细点,谢谢!
(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
1/x - 1/(x+1) 通分得到1/x(x+1)
所以1/x(x+1)=1/x - 1/(x+1)
所以1/2*3=1/2-1/3
1/3*4=1/3-1/4
……
(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 (中间项消去)
=99/100
此为裂项求和法,1/n(n+1)=1/n-1/(n+1),所以
原式等于1-1/2+1/2-1/3+1/3-1/4+……+1/99-1/100
=1-1/100=99/100
因为1/1*2=1-1/2;1/2*3=1/2-1/3;依次类推,1/99*100=1/99-1/100.
所以原题等于1-1/2+1/2-1/3+1/3-1/4........+1/99-1/100=1-1/100=99/100
1/2-1/2=?
1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+100)
1+1/(1+2)+1/(1+2+3)+-------+1/(1+2+3+----+100)
1+1/1+2+1/1+2+3+...+1/1+2+3...+2000
1+1/1+2+1/1+2+3.........+1/1+2+3.....100
1*(1/1+2)*(1/1+2+3)*~~~*(1/1+2+~~~2005)=?
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)
(1-1/2^2)*(1-1/3^2)*(1-1/4^2).......(1-1/100^2)
3/2=2+1/1*2=1/1+1/2
(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1)