计算1⼀1+2+1⼀1+2+3+.....1⼀1+2+3......100

2024-12-20 06:16:16
推荐回答(2个)
回答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

回答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