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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]

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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. \(\frac{(m-1)m}{2} - \frac{n(n+1)}{2}\)
[Reveal] Spoiler: OA

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Re: The sum of the first k positive integers is equal to k(k+1)/2. What is [#permalink]

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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]

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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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Re: The sum of the first k positive integers is equal to k(k+1)/2. What is [#permalink]

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New post 19 Jul 2015, 05:05
Can also be solved by using Sum of A.P formula from n to m. (A bit lengthy though)

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Re: The sum of the first k positive integers is equal to k(k+1)/2. What is [#permalink]

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New post 16 Sep 2015, 20:52
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

I got Answer B.
But OA is different.


I think this question can be solved easily by picking numbers.

Let n = 1 and m = 2

Sum of 1 integer is 1;
Sum of 2 integers is 3

So, Sum of the integers from 1 to 2 must be 3. Let's pluck N and M in the choices

A. \(\frac{2(2+1)}{2}\) - \(\frac{(1+1)(1+2)}{2}\) \(= 3 - 3 = 0\)

B. \(\frac{2(2+1)}{2}\) - \(\frac{1(1+1)}{2}\) \(= 3 - 1 = 2\)

C. \(\frac{2(2+1)}{2}\) - \(\frac{(1-1)1}{2}\) \(= 3 - 0 = 3\) Bingo!

D. \(\frac{(2-1)2}{2}\) - \(\frac{(1+1)(1+2)}{2}\) \(= 1 - 3 = -2\)

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

Correct me if I'm wrong pls

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Re: The sum of the first k positive integers is equal to k(k+1)/2. What is [#permalink]

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New post 20 Dec 2015, 17:08
I solved by picking numbers.
n=12
m=15.

only answer choice C yields a valid result.

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Re: The sum of the first k positive integers is equal to k(k+1)/2. What is [#permalink]

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New post 20 Dec 2015, 21:14
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]

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