[(1+2)/2][(1+2+3)/(2+3)][(1+2+3+4)/(2+3+4)]......[(1+2+3+......+50)/(2+3+......+50)]

来源：百度知道编辑：UC知道时间：2024/09/21 17:35:55

(1+2+3+......+n)/(2+3+......+n)
=[n(n+1)/2]/[n(n+1)/2-1]
=(n^2+n)/(n^2+n-2)
=n(n+1)/(n+2)(n-1)

所以[(1+2)/2]*[(1+2+3)/(2+3)]*[(1+2+3+4)/(2+3+4)]*......*[(1+2+3+......+50)/(2+3+......+50)]
=(2*3/1*4)*(3*4/2*5)(4*5/3*6)......(49*50/48*51)(50*51/49*52)
=3*50/52
=75/26

[(1+2)/2]*[(1+2+3)/(2+3)]*[(1+2+3+4)/(2+3+4)]*......*[(1+2+3+......+50)/(2+3+......+50)]
=[2*3/4*1]*[3*4/5*2]*[4*5/6*3]*......*[50*51/52*49]
=3*50/52
=150/52
=75/26

1/2-1/2=? (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 1/1^2+1/2^2+...+1/n^2<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)=?