Baten80 wrote:

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

I got Answer B.

But OA is different.

I think this question can be solved easily by picking numbers.

Let n = 1 and m = 2

Sum of 1 integer is 1;

Sum of 2 integers is 3

So, Sum of the integers from 1 to 2 must be 3. Let's pluck N and M in the choices

A. \(\frac{2(2+1)}{2}\) - \(\frac{(1+1)(1+2)}{2}\) \(= 3 - 3 = 0\)

B. \(\frac{2(2+1)}{2}\) - \(\frac{1(1+1)}{2}\) \(= 3 - 1 = 2\)

C. \(\frac{2(2+1)}{2}\) - \(\frac{(1-1)1}{2}\) \(= 3 - 0 = 3\)

Bingo!D. \(\frac{(2-1)2}{2}\) - \(\frac{(1+1)(1+2)}{2}\) \(= 1 - 3 = -2\)

E. \(\frac{(2-1)2}{2}\) - \(\frac{1(1+1)}{2}\) \(= 1 - 1 = 0\)

Correct me if I'm wrong pls