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