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

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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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02 Oct 2018, 01:21
1
Bunuel wrote:
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: $$\frac{m(m+1)}{2}-\frac{(n-1)(n-1+1)}{2}=\frac{m(m+1)}{2}-\frac{(n-1)n}{2}$$.

Hi Bunuel. I did not follow the part were you took (n-1). Can you please explain the same. Thanks.
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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 Aug 2019, 03:36
1
09173140521 wrote:
I GO FOR B but wrong answer .....why we must consider (n-1) although I consider just (n)? cannot find what logic behind that

I seem to have an answer to my own question. If you read the question stem closely then it asks for sum from $$n$$ to $$m$$, inclusive. The word inclusive is the key here. If you take $$n$$ then you make the sum exclusive of $$n$$, but if you take $$(n-1)$$ then you make the sum inclusive.

Bunuel - Please correct if my understanding is incorrect. Thanks for the great explanation in your initial post!
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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 Aug 2019, 03:45
1
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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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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11 Aug 2019, 03:20
Pritishd wrote:
Bunuel wrote:
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: $$\frac{m(m+1)}{2}-\frac{(n-1)(n-1+1)}{2}=\frac{m(m+1)}{2}-\frac{(n-1)n}{2}$$.

Hi Bunuel. I did not follow the part were you took (n-1). Can you please explain the same. Thanks.

good question maybe someday Bunuel answer that
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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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11 Aug 2019, 03:24
09173140521 wrote:
Pritishd wrote:
Bunuel 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: $$\frac{m(m+1)}{2}-\frac{(n-1)(n-1+1)}{2}=\frac{m(m+1)}{2}-\frac{(n-1)n}{2}$$.

Hi Bunuel. I did not follow the part were you took (n-1). Can you please explain the same. Thanks.

good question maybe someday Bunuel answer that

_________________________
Not sure what to explain there.
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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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11 Aug 2019, 03:34
I GO FOR B but wrong answer .....why we must consider (n-1) although I consider just (n)? cannot find what logic behind that
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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 Aug 2019, 03:43
Pritishd wrote:
09173140521 wrote:
I GO FOR B but wrong answer .....why we must consider (n-1) although I consider just (n)? cannot find what logic behind that

I seem to have an answer to my own question. If you read the question stem closely then it asks for sum from $$n$$ to $$m$$, inclusive. The word inclusive is the key here. If you take $$n$$ then you make the sum exclusive of $$n$$, but if you take $$(n-1)$$ then you make the sum inclusive.

Bunuel - Please correct if my understanding is incorrect. Thanks for the great explanation in your initial post!
yes it is correct ! ...the key is <inclusive> ....many thanks
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Re: The sum of the first k positive integers is equal to k(k+1)/2. What is   [#permalink] 16 Aug 2019, 03:43
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