求数列1,1/(1+2),1/(1+2+3)……1/(1+2+…n)……前n项的和

来源:百度知道 编辑:UC知道 时间:2024/09/24 17:08:47

1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+......+1/(1+2+3+...+n)
=1+ 2/2*3+2/3*4+2/4*5+......+2/n(n+1)
=1+2(1/2-1/3+1/3-1/4+...+1/n-1/(n+1))
=1+2[1/2-1/(n+1)]
=2-2/(n+1)
=2n/(n+1)

当n趋于无穷大时 (n-1)/(n+1) =1,即Xn=1+1/(1+2)+1/(1+2+3)+……+1/(1+2+……+n)为2

原式=1+1/3+1/6+……+1/((n+1)*n/2)
原式=1/2+1/6+1/12+……+1/((n+1)*n)
=1-1/2+1/2-1/3+1/3-1/4+……+1/n-1/(n+1)
=1-1/(n+1)=n/(n+1)

1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+......+1/(1+2+3+...+n)
=1+ 2/2*3+2/3*4+2/4*5+......+2/n(n+1)
=1+2(1/2-1/3+1/3-1/4+...+1/n-1/(n+1))
=1+2[1/2-1/(n+1)]
=2-2/(n+1)
=2n/(n+1)