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For any positive integer n, the sum of the first n positive

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For any positive integer n, the sum of the first n positive  [#permalink]

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New post 27 Apr 2014, 21:00
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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
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Re: For any positive integer n, the sum of the first n positive  [#permalink]

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New post 28 Apr 2014, 01:25
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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


The sum of all the integers between m and n, inclusive is the sum of the first m positive integers minus the sum of the first n-1 positive integers.

The sum of the first m positive integers is m(m+1)/2;

The sum of the first n-1 positive integers is (n-1)n/2;

Therefore the answer is m(m+1)/2 - (n-1)n/2.

Answer: B.
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Re: For any positive integer n, the sum of the first n positive  [#permalink]

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New post 28 Apr 2014, 01:19
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 here

Take 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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Re: For any positive integer n, the sum of the first n positive  [#permalink]

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New post 05 May 2014, 23:04
To both Experts: Thank you very much, helpful as usual
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Re: For any positive integer n, the sum of the first n positive  [#permalink]

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New post 13 Mar 2016, 07:54
Hi Bunuel,

Plugging in number that's what helped me personally, but could you explain why general formula n(n+1)/2 doesn't work here? Is it because we include m and and and need to subtract these two terms? Than why do we do this exactly with the second part of the equation, not subtracting 1 from the 1st one and 1 from the second?
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Re: For any positive integer n, the sum of the first n positive  [#permalink]

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New post 16 Mar 2016, 18:37
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


sum of the first n numbers: n(n+1)/2
sum of the first m numbers: m(m+1)/2

sum of the integers between m and n - excluding m and n:
m(m+1)/2 - n(n+1)/2
we can eliminate A and E right away, because the sign is +
C - is the sum of the integers between m and n, but without m and n so out.
E - clearly not correct, as it mixes m with n...

B
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Re: For any positive integer n, the sum of the first n positive  [#permalink]

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New post 18 Mar 2016, 12:41
Can somebody please elaborate on how B is the option.Is the sum of integers between m and not equal to difference between sum of first m integers and sum of first n integers.I am not clear with this.Please explain
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Re: For any positive integer n, the sum of the first n positive  [#permalink]

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New post 16 Aug 2017, 06:44
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


Missed this simple question because of over-confidence. Actually, it was more like - 'Is this really a GMAT type question?'

Anyhow, lesson - if you finish solving a question super quick, you need to check where the tricky part is.
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Re: For any positive integer n, the sum of the first n positive  [#permalink]

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New post 09 Mar 2019, 15:34
bhamini1 wrote:
Can somebody please elaborate on how B is the option.Is the sum of integers between m and not equal to difference between sum of first m integers and sum of first n integers.I am not clear with this.Please explain


I was confused about this as well, I think this is how it works:

If we're looking at the number of terms between 7 and 4, inclusive, we would use First - Last + 1 --> 7 - 4 + 1 = 4
Why do we add 1 here? Because we need to include the 4 (the terms are 7,6,5 AND 4).

It's the same idea here. Let's take m = 7 and n = 4, we're looking for the sum of first 7 ints - first 4 int:

7+6+5+4+3+2+1
-______4+3+2+1

In the problem, it says inclusive, so we must include the first and last ints in the set. In this case, we have to subtract 1 from n because otherwise we will remove the 4 (and thus part of the set that we need):

7+6+5+4+3+2+1
-________3+2+1

If we use n, n+1 we would actually end up removing even more of the terms. The idea is that we're looking for a chunk of sums between the biggest number and the smallest number, but we can't remove the smallest number so we have to make sure we subtract 1 less than that smallest number.
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Re: For any positive integer n, the sum of the first n positive   [#permalink] 09 Mar 2019, 15:34
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