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# The sum of the first k positive integers is equal to k(k+1)/

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The sum of the first k positive integers is equal to k(k+1)/ [#permalink]  19 Jan 2012, 10:16
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45% (02:42) correct 55% (01:20) wrong based on 81 sessions
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

But OA is different.
[Reveal] Spoiler: OA

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Re: The sum of the first k positive integers is equal to k(k+1)/ [#permalink]  19 Jan 2012, 10:28
Expert's post
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Baten80 wrote:
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

But OA is different.

The sum of the integers from n to m, inclusive, will be the sum of the first m positive integers minus the sum of the first n-1 integers: $$\frac{m(m+1)}{2}-\frac{(n-1)(n-1+1)}{2}=\frac{m(m+1)}{2}-\frac{(n-1)n}{2}$$.

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Re: The sum of the first k positive integers is equal to k(k+1)/ [#permalink]  19 Jan 2012, 10:34
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Expert's post
Baten80 wrote:
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

But OA is different.

Or try plug-in method: let m=4 and n=3 --> then m+n=7. Let see which option yields 7.
A. m(m+1)/2 - (n+1)(n+2)/2 = 10-10=0;
B. m(m+1)/2 - n(n+1)/2 = 10-6=4;
C. m(m+1)/2 - (n-1)n/2 = 10-3=7 --> OK;
D. (m-1)m/2 - (n+1)(n+2)/2 = 6-10=-4;
E. (m-1)m/2 - n(n+1)/2 = 6-6=0.

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Re: The sum of the first k positive integers is equal to k(k+1)/ [#permalink]  03 Jul 2014, 01:09
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: The sum of the first k positive integers is equal to k(k+1)/   [#permalink] 03 Jul 2014, 01:09
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