计算1/1+2+1/1+2+3+.....1/1+2+3......100

Answer 1

1/1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+…1/(1+2+3+4+5+…n)

=2{1/(1×2）+1/（2×3）+…+1/[n(n+1)]}

=2[1-1/2+1/2-1/3+…+1/n-1/(n+1)]

=2[1-1/(n+1)]

=2n/(n+1)

将100代入得到
式子=200/101

追问

1/(1+2)+1/(1+2+3)+.....1/(1+2+3....100).

追答

1+2=2*3/2
1+2+3=3*4/2
......
1+2+3+4+......+100=100*101/2
所以
1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+4+......+100)
=2(1/2 -1/3) +2(1/3-1/4)+......+2(1/100-1/101)
=2*(1/2-1/101)
=99/101

Answer 2

she f(n)=1+2+.......+n
即f（n）=(1+n)/2
由题知道s=f(1)+f(2)+....f(n)-1
s=(n/2+1/2+2/2+......n/2)-1
s=n/2+(1+n)/4=(3n-3)/4
希望对你有所帮助
s(100)=297/4