1/1+2 + 1/1+2+3 + 1/1+2+3+4 + 1/1+2+3+4+…+99
来源:百度知道 编辑:UC知道 时间:2024/06/05 08:11:52
应用通项公式1+2+...+n=n(n+1)/2
原式=2/(2*3)+2/(3*4)+2/(4*5)+...+2/(99*100)
=2*(1/2-1/3+1/3-1/4+1/4-1/5+...+1/99-1/100)
=2*(1/2-1/100)
=49/50
因为1+2+3+4+。。。。。+n可以表示为n*(n+1)/2,能看懂不?
1+2+3+4+5+。。。+99=99*(99+1)/2
1/[n*(n+1)/2]=2/[n(n+1)]=2[1/n-1/(n+1)]
则,题目里的式子=2[(1/2)-(1/3)+(1/3)-(1/4)+(1/4)-(1/5)+(1/5)-(1/6)+...........+(1/99)-(1/100)]=2[(1/2)-(1/100)]=49/50
设n为项数(1-98项)
1/[(1+n+1)*(n+1)/2]=2/[(n+1)*(n+2)]=2*(1/(n+1)-1/(n+2))
合并
2*[1/2-1/3+1/3-1/4.......+1/99-1/100]=2*[1/2-1/100]
=49/50
(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1)
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)(1-1/3)(1-1/4)(1-1/5).....(1-1/1000)
(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)
1+1/2+1+1/3+1+1/4+......+1/100=?