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The sum of the first k positive integers is equal to k(k+1)/2. What is

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The sum of the first k positive integers is equal to k(k+1)/2. What is [#permalink] New post   Updated on: 05 Dec 2020, 09:17
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The sum of the first k positive integers is equal to \(\frac{k(k+1)}{2}\). What is the sum of the integers from n to m, inclusive, where 0<n<m?

A. \(\frac{m(m+1)}{2} - \frac{(n+1)(n+2)}{2}\)

B. \(\frac{m(m+1)}{2} - \frac{n(n+1)}{2}\)

C. \(\frac{m(m+1)}{2} - \frac{(n-1)n}{2}\)

D. \(\frac{(m-1)m}{2} - \frac{(n+1)(n+2)}{2}\)

E. \(\frac{(m-1)m}{2} - \frac{n(n+1)}{2}\)
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C

Originally posted by MHIKER on 19 Jan 2012, 11:16.
Last edited by MHIKER on 05 Dec 2020, 09:17, edited 1 time in total.
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Re: The sum of the first k positive integers is equal to k(k+1)/2. What is [#permalink] New post   19 Jan 2012, 11:28
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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. \(\frac{m(m+1)}{2} - \frac{(n+1)(n+2)}{2}\)

B. \(\frac{m(m+1)}{2} - \frac{n(n+1)}{2}\)

C. \(\frac{m(m+1)}{2} - \frac{(n-1)n}{2}\)

D. \(\frac{(m-1)m}{2} - \frac{(n+1)(n+2)}{2}\)

E. \(\frac{(m-1)m}{2} - \frac{n(n+1)}{2}\)


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: \(\frac{m(m+1)}{2}-\frac{(n-1)(n-1+1)}{2}=\frac{m(m+1)}{2}-\frac{(n-1)n}{2}\).

Answer: C.
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Re: The sum of the first k positive integers is equal to k(k+1)/2. What is [#permalink] New post   19 Jan 2012, 11:34
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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. \(\frac{m(m+1)}{2} - \frac{(n+1)(n+2)}{2}\)

B. \(\frac{m(m+1)}{2} - \frac{n(n+1)}{2}\)

C. \(\frac{m(m+1)}{2} - \frac{(n-1)n}{2}\)

D. \(\frac{(m-1)m}{2} - \frac{(n+1)(n+2)}{2}\)

E. \(\frac{(m-1)m}{2} - \frac{n(n+1)}{2}\)



Or try plug-in method: let m=4 and n=3 --> then m+n=7. Let see which option yields 7.

A. \(\frac{m(m+1)}{2} - \frac{(n+1)(n+2)}{2} = 10-10=0\);

B. \(\frac{m(m+1)}{2} - \frac{n(n+1)}{2} = 10-6=4\);

C. \(\frac{m(m+1)}{2} - \frac{(n-1)n}{2} = 10-3=7\) --> OK;

D. \(\frac{(m-1)m}{2} - \frac{(n+1)(n+2)}{2} = 6-10=-4\);

E. \(\frac{(m-1)m}{2} - \frac{n(n+1)}{2} = 6-6=0\).


Answer: C.
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neha1993
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Re: The sum of the first k positive integers is equal to k(k+1)/2. What is [#permalink] New post   20 Dec 2015, 22:14
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The only thing to trick here is that we need the sum of (n-1) integers to be subtracted from the sum of m integers.
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Re: The sum of the first k positive integers is equal to k(k+1)/2. What is [#permalink] New post   16 Aug 2019, 04:45
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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. \(\frac{m(m+1)}{2} - \frac{(n+1)(n+2)}{2}\)

B. \(\frac{m(m+1)}{2} - \frac{n(n+1)}{2}\)

C. \(\frac{m(m+1)}{2} - \frac{(n-1)n}{2}\)

D. \(\frac{(m-1)m}{2} - \frac{(n+1)(n+2)}{2}\)

E. \(\frac{(m-1)m}{2} - \frac{n(n+1)}{2}\)


Given: The sum of the first k positive integers is equal to k(k+1)/2.

Asked: What is the sum of the integers from n to m, inclusive, where 0<n<m?

Sum of the integers from n to m, inclusive, where 0<n<m = Sum of first m integers - Sum of first (n-1) integers
= \(\frac{m(m+1)}{2} - \frac{(n-1)n}{2}\)

IMO C
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The sum of the first k positive integers is equal to k(k+1)/2. What is [#permalink] New post   25 Jan 2024, 09:18
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Dhruv1212 wrote:
can someone explain again why it was n-1


We need the sum of the integers from n to m, inclusive. So, we need to subtract the sum of the integers from 1 to (n-1), inclusive, from the sum of the integers from 1 to m, inclusive, to get the sum of the integers from n to m, inclusive. For example, to get the sum of the integers from 5 to 10, inclusive (5+6+7+8+9+10), we subtract the sum of the integers from 1 to 4 (1+2+3+4) from the sum of the integers from 1 to 10 (1+2+3+...+10).

Hope it's clear.
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Re: The sum of the first k positive integers is equal to k(k+1)/2. What is [#permalink] New post   25 Jan 2024, 05:47
can someone explain again why it was n-1
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Re: The sum of the first k positive integers is equal to k(k+1)/2. What is [#permalink] New post   02 Feb 2024, 15:46
What helped me understand why its (n-1) is set up the equation starting with
(m(m+1)/2) minus (n(n+1)/2)
Then, I realized that n has to be subtracted from the n(n+1)/2 so that it will technically be counted as part of the m(m+1)/2 which took me a while to think about. Just have to realize that for n to be included in m then we have to subtract it out of what will be subtracted... if that makes sense.

So we have (n(n+1)/2)- n

When simplifying we get ((n^2+n)/2)-(2n/2) which simplifies to (n^2+n-2n)/2 which simplifies to n(n-1)/2

Hope that helps
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Re: The sum of the first k positive integers is equal to k(k+1)/2. What is [#permalink]
02 Feb 2024, 15:46
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