Question

已知数列1,1/1+2,1/1+2+3,…,1/1+2+3+…+n,…，则其前n项的和等于多少?

Answer 1

1＝2/2＝2/(1×2)＝2(1－1/2)
1/(1＋2)＝2/6＝2/(2×3)＝2(1/2－1/3)
1/(1＋2＋3)＝2/12＝2/(3×4)＝2(1/3－1/4)
…
1/(1＋2＋3＋…＋n)＝1/[n(n＋1)/2]＝2/[n(n＋1)]＝2[1/n－1/(n＋1)]
故前n项的和＝2(1－1/2)＋2(1/2－1/3)＋2(1/3－1/4)＋…＋2[1/n－1/(n＋1)]
＝2[1－1/2＋1/2－1/3＋1/3－1/4＋…＋1/n－1/(n＋1)]
＝2[1－1/(n＋1)]
＝2n/(n＋1)

Answer 2

1+2+3+...+n
=n(n+1)/2

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

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

Answer 3

(2N)/(N+1)

Answer 4

2n/(n+1)