1*2/1+2*3/1+3*4/1+4*5/1+......99*100/1
来源:百度知道 编辑:UC知道 时间:2024/06/18 17:42:48
求救!有会的请尽快帮忙
原式=(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/99-1/100)
中间正负抵消
=1/1-1/100
=99/100
1/n(n+1)=1/n-1/(n+1)
1*2/1+2*3/1+3*4/1+4*5/1+......99*100/1
=1-(1/2)+(1/2)-(1/3)+(1/3)-(1/4)+(1/4)-(1/5)+……+(1/99)-(1/100)
=1-(1/100)
=99/100
这是裂项法 1*2/1=1/1-1/2 2*3/1=1/2-1/3).....以此类推...........中间抵消
得:
1/1-1/100 =99/100
答:1*2/1+2*3/1+3*4/1+4*5/1+......99*100/1=99/100.
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+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=
3/2=2+1/1*2=1/1+1/2
(1-1/2^2)*(1-1/3^2)*(1-1/4^2).......(1-1/100^2)
1+1/2+1/3+.....+1/n
1+1/2+1/3+...+1/100
1-1/2+1/3-.....-1/10