1/(2*3)+1/(3*4)+1/(4*5)…+1/(99*100)=
来源:百度知道 编辑:UC知道 时间:2024/07/03 10:24:08
1/(2*3)+1/(3*4)+1/(4*5)…+1/(99*100)=
一定要简便方法啊!而且要有过程
一定要简便方法啊!而且要有过程
1/(2*3)+1/(3*4)+1/(4*5)…+1/(99*100)
=1/2-1/3+1/3-1/4+......+1/99-1/100
=1/2-1/100
=49/100
=(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+...(1/99-1/100)
=1/2-1/3+1/3+1/4-1/4+...+1/99-1/100
=1/2-1/100
=49/100
=1/2-1/3+1/3-1/4+1/4-1/5+.....+1/99-1/100
=1/2-1/100
=49/100
您好,我的回答是这样的:
1/(2*3)+1/(3*4)+1/(4*5)…+1/(99*100)
=(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+...(1/99-1/100)
=1/2-1/3+1/3+1/4-1/4+...+1/99-1/100
=1/2-1/100
=49/100
如上解答都正确 这种类型的一般通用裂相法将中间的都抵消了 就剩下首相和尾相
1/(n*(n+1))=1/n-1/(n+1)
∴1/(2*3)+1/(3*4)+1/(4*5)…+1/(99*100)
=(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+...(1/99- 1/100)
=1/2-1/3+1/3-1/4+......+1/99-1/100
=1/2-1/100
=49/100
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+----+100)
(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1)
1+1/2+1/3+.....+1/n
1+1/2+1/3+...+1/100
1-1/2+1/3-.....-1/10
(1+1/2+1/3+1/4)×
1+1/1+2+1/1+2+3+...+1/1+2+3...+2000
1*(1/1+2)*(1/1+2+3)*~~~*(1/1+2+~~~2005)=?
1/1+2 + 1/1+2+3 +....+ 1/1+2+3+....+100=