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
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