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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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19 Jan 2012, 10:16
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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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19 Jan 2012, 10:28
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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: $$\frac{m(m+1)}{2}-\frac{(n-1)(n-1+1)}{2}=\frac{m(m+1)}{2}-\frac{(n-1)n}{2}$$.

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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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19 Jan 2012, 10:34
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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. $$\frac{(m-1)m}{2} - \frac{n(n+1)}{2} = 6-6=0$$.

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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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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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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

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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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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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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01 Jan 2017, 13:00
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Re: The sum of the first k positive integers is equal to k(k+1)/2. What is   [#permalink] 01 Jan 2017, 13:00
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