1+1⼀(1+2)+1⼀(1+2+3)+1⼀(1+2+3+4)+......+1⼀(1+2+3+...+100)

简算
2024-12-28 07:19:20
推荐回答(1个)
回答1:

1+2+3+...+n=n(n+1)/2
1/(1+2+3+...+n)=2/n(n+1)=2[1/n-1/(n+1)]
1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+......+1/(1+2+3+...+100)
=2[(1-1/2)+(1/2-1/3)+(1/3-1/4)+......+(1/100-1/101)]
=2(1-1/101)
=200/101