题目:
The sum of the first k positive integers is equal to $$frac{k(k+1)}{2}$$. What is the sum of the integers from n to m, inclusive, where $$0<{n}<{m}$$?
选项:
A、 $$\frac{m(m+1)}{2}-\frac{({n}+{1})({n}+{2})}{2}$$
B、 $$\frac{m(m+1)}{2}-\frac{(n-1)n}{2}$$
C、 $$\frac{m(m+1)}{2}-\frac{(n+1)(n+2)}{2}$$
D、 $$\frac{(m-1)m}{2}-\frac{(n+1)(n+2)}{2}$$
E、 $$\frac{(m-1)m}{2}-\frac{n(n+1)}{2}$$
答案:
B