Question

1+2分之1+1+2+3分之1+1+2+3+4分之1+.+1+2+3+.+n分之1,求它们的和

Answer 1

1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+……+1/(1+2+3……+n)
= 1+1/[(1+2)×2÷2]+1/[(1+3)×3÷2]+……+1/[(1+n)×n÷2]——①
= 2/2+2/(1+2)×2+2/(1+3)×3+……+2/(1+n)×n——②
= 2×[1/2+1/2-1/3+1/3-1/4+……+1/n-1/(1+n)]——③
= 2×[1-1/(1+n)]
= 2×[n/(1+n)]
= 2n/(1+n)
注释：
①把分母等差数列写成简便形式
②分子和分母同时乘以2
③把分子2提出来做公因数