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

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为100，∴带入即可得到200/101

Answer 2

1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+……+1/(1+2+3……+100)
= 1+1/[(1+2)×2÷2]+1/[(1+3)×3÷2]+……+1/[(1+100)×100÷2]
= 2/2+2/(1+2)×2+2/(1+3)×3+……+2/(1+100)×100
= 2×[1/2+1/2-1/3+1/3-1/4+……+1/100-1/(1+100)]
= 2×[1-1/(1+100)]
= 2×[100/(1+100)]
= 2100/(1+100)