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?

A. m(m+1)/2-(n+1)(n+2)/2

B. m(m+1)/2-n(n+1)/2

C. m(m+1)/2-(n-1)n/2

D. (m-1)m/2-(n+1)(n+2)/2

E. (m-1)m/2-n(n+1)/2

The answer is C

Please help to explain it

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

lz一定是忘了应该加上n，不过我第一次做的话也很有可能会忘的