i2014 wrote:

For any positive integer n, the sum of the first n positive integers equals n(n+1)/2 . If m and n are positive integers and m > n, what is the sum of all the integers between m and n, inclusive?

(A) m(m+1)/2 + n(n+1)/2

(B) m(m+1)/2 - n(n-1)/2

(C) m(m+1)/2 - n(n+1)/2

(D) m(m+1)/2 + n(n-1)/2

(E) m(n+1)/2 - n(m-1)/2

Sol:

Plug in is the best way forward hereTake m=5 n=3 and we get

Sum of first 5 terms = (5*6)/2 =15

Sum of first 3 terms (n) = (3*4)/2= 6

sum of all the integers between m and n, inclusive: 3+4+5= 12

Starting with option and plugging the values of m and n, we see that option B gives us 12.

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