[讨论]GWD-21-Q2

The sum of the first k positive integers is equal to, k(k+1)/2, What is the sum of the integers from n to m, inclusive, where 0<n<m?

gwd 答案为 C m(m＋1)/2  -（n－1)n/2

请求思路！

m(m+1)/2-n(n+1)/2+n=m(m+1)/2-n(n-1)/2

因为n和m是inclusive的，而直接减的话会把n减掉，所以要重新加回来

Thanks a lot!got that!

I feel i am a fool.....