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If the sum of the first k positive integers is equal to, k(k

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by Max@Math Revolution » Mon Jan 25, 2016 4:41 pm
If 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)-(n-1)n)/2
B. (m(m+1)-(n+1)n)/2
C. (m(m-1)-(n-1)n)/2
D. (m(m-1)-(n+1)n)/2
E. (m(m+1)-(n-1)n)/4


* A solution will be posted in two days.
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by Testtrainer » Tue Jan 26, 2016 10:46 pm
Pick m = 4, n = 2.

Sum of 2 + 3 + 4 = 9

(A) 9
(B) 7
(C) 5
(D) 3
(E) 9/2

Variables in the answers: always think about picking numbers, staying away from zero and one, and you should be good to go.
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by Max@Math Revolution » Wed Jan 27, 2016 5:38 pm
If 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)-(n-1)n)/2
B. (m(m+1)-(n+1)n)/2
C. (m(m-1)-(n-1)n)/2
D. (m(m-1)-(n+1)n)/2
E. (m(m+1)-(n-1)n)/4


==> n+(n+1)+.......+(m-1)+m={1+2+...+(n-1)+n+(n+1)+...+(m-1)+m}-{1+2+...+(n-1)}=m(m+1)/2-n(n-1)/2.

Therefore, the answer is A.
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