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}?
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A\frac{m(m+1)}{2}-\frac{({n}+{1})({n}+{2})}{2}
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B\frac{m(m+1)}{2}-\frac{(n-1)n}{2}
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C\frac{m(m+1)}{2}-\frac{n(n+1)}{2}
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D\frac{(m-1)m}{2}-\frac{(n+1)(n+2)}{2}
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E\frac{(m-1)m}{2}-\frac{n(n+1)}{2}
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正确答案: B