Question

奥数题1+1/1+2+1/1+2+3+1/1+2+3+4+……+1/1+2+3+4+5+……100

Answer 1

推导过程：
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)
此题中n=100，带入即可得到200/101

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

Answer 2

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]
= 2/2+2/(1+2)×2+2/(1+3)×3+……+2/(1+n)×n
= 2×[1/2+1/2-1/3+1/3-1/4+……+1/n-1/(1+n)]
= 2×[1-1/(1+n)]
= 2×[n/(1+n)]
= 2n/(1+n)
此题中n=100，带入即可得到200/101

Answer 3

原式=1+（1+2）+（1+2+3）+。。。。。。+（1+2+3+。。。+n）（n=100）
令an=1+2+3+。。。+n=n（n+1）/2
原式=sn=a1+a2+。。。。+an=n（n+1）/4+n（n+1）（2n+1）/6=340875（n=100）

Answer 4

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)
此题中n=100，带入即可得到200/101

选我哦

Answer 5

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)
此题中n=100，带入即可得到200/101