Question

1/1+2,1/1+2+3,1/1+2+3+4,...,1/1+2+3+...+n的和

Answer 1

1+2+3+……+n=n(n+1)/2
1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+,...,+1/(1+2+3+...+n)
=2/2×3+2/3×4+2/4×5+……+2/n(n+1)
=2(1/2×3+1/3×4+1/4×5+……+1/n(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

数列通项an=1/[n(n+1)/2]
=2[1/n-1/(n+1)];
分别令n=1,2,3, ..., n,得n个式子;
将这n个式子两边相加即得,
前n项和 Sn=[1/1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+...+n)]-1
=2×[1-1/(n+1)]-1
=(n-1)/(n+1)。

追问

我还没学数列通项 但是我大概了解了一下 可是不知道怎么做