MathRevolution wrote:
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.
If you don't know how to manipulate that particular formula, or if the manipulation takes too long, choose values.
Let n = 2 and m = 4
Find the sum of the integers from n to m, inclusive. Either
1) 2 + 3 + 4 = 9, or
2) [IF you choose numbers too far apart]: (average) * (# of terms) is
\(\frac{(2+4)}{2}\) * 3 = 9
Check answer choices with n = 2, m = 4. The one that yields 9 as the answer is correct.
The arithmetic with these numbers is quick.
Answer A. (m(m+1)-(n-1)n)/2
(4*5 - 1*2)/2 = 18/2 = 9. MATCH
I checked the others, quickly.
B. (m(m+1)-(n+1)n)/2
(20 - 6)/2 = 7. NOT a match
C. (m(m-1)-(n-1)n)/2
(12 - 1) = 11 won't work as a numerator. Move on. NOT a match
D. (m(m-1)-(n+1)n)/2
(12 - 6) = 6, already too small without division. Move on. NOT a match.
E. (m(m+1)-(n-1)n)/4
(20 - 1) = 19 won't work as a numerator. Move on. NOT a match
Answer A _________________
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