Question

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

Answer 1

先由分母可得分母的同项公式是n(n+1)/2
所以Sn=2(1/2+1/6+1/12.....+1/n(n+1)
=2[1-1/2+1/2-1/咐和3+1/3-1/帆含4....+1/n-1/(n+1)]
注:1/2=1-1/2 1/6=1/2-1/态简笑3 ...
1/n(n+1)=1/n-1/(n+1)
可得Sn=2[1-1/(n+1)]
=2n/(n+1)

Answer 2

Sn=1+1/(1+2)+1/(1+2+3)+.....+1/(1+2+3+....+n)
=1+2/罩悄掘2*(1+2)+2/3*(3+1)+..+2/n(n+1)
=2(1/2+1/2-1/3+1/3-1/4+..+1/物核n-1/(n+1)
=2(1-1/运雀(n+1))
=2n/(n+1)

Answer 3

原式=1/[(1+1)*1/橘铅2]+1/[(1+2)*2/2]+……+1/[(1+n)*n/2]
=2/(1*2)+2/(2*3)+2/(3*4)+…圆哪好…+2/[n(n+1)]
=2(1-1/2+1/2-1/3+……+1/n-1/(n+1))
=2*(1-1/缓枝(n+1))

Answer 4

运用 1+2+...+k = k(k+1)/2
和 1/(k(k+1)) = (k+1-k)/(k(k+1)) = 1/k - 1/(k+1)

Sn=1+1/(1+2)+1/(1+2+3)+.....+1/(1+2+3+....+n)
= 2/(1*2) + 2/迟档者蠢嫌(2*3) + 2/(3*4) + 2/(4*5) + ... + 2/(n(n+1))
= 2[(1/1 - 1/2) + (1/码薯2 - 1/3) + (1/3 - 1/4) + (1/n - 1/(n+1))]
= 2[1/1 - 1/(n+1)]
= 2n/(n+1)

Answer 5

通项是
an=1/春让(1+2+3....+n)=2/n(n+1)=2*[1/n-1/(n+1)]
所以
Sn=1-1/（n+1）=n/（扒态局n+1）闭雀