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If the sum of the first n positive integers is S, what is

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If the sum of the first n positive integers is S, what is [#permalink]

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If the sum of the first n positive integers is S, what is the sum of the first n positive even integers, in terms of S ?

(A) S/2
(B) S
(C) 2S
(D) 2S + 2
(E) 4S
[Reveal] Spoiler: OA

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Re: p/s sum of integers [#permalink]

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New post 08 Mar 2011, 06:24
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Let n = 6.
S = n(n+1)/2 = 3 * 7 = 21
Sum of first 6 even = 2 + 4 + 6 + 8 + 10 + 12 = 42 = 2S

Hence C.

Lolaergasheva wrote:
If the sum of the first n positive integers is S, what is the sum of the first n positive even integers, in terms of S ?
(A) S/2
(B) S
(C) 2S
(D) 2S + 2
(E) 4S

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Re: p/s sum of integers [#permalink]

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New post 08 Mar 2011, 06:32
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Lolaergasheva wrote:
If the sum of the first n positive integers is S, what is the sum of the first n positive even integers, in terms of S ?
(A) S/2
(B) S
(C) 2S
(D) 2S + 2
(E) 4S


1+2+3+...+n=S
2+4+6+...+2n=2(1+2+3+...+n)=2S.

Or you can simply take n=2 --> 1+2=3=S --> 2+4=6=2S.

Answer: C.

Generally:
Sum of n first integers: \(1+2+...+n=\frac{1+n}{2}*n\)

Sum of n first odd numbers: \(a_1+a_2+...+a_n=1+3+...+a_n=n^2\), where \(a_n\) is the last, \(n_{th}\) term and given by: \(a_n=2n-1\). Given \(n=5\) first odd integers, then their sum equals to \(1+3+5+7+9=5^2=25\).

Sum of n first positive even numbers: \(a_1+a_2+...+a_n=2+4+...+a_n\)\(=n(n+1)\), where \(a_n\) is the last, \(n_{th}\) term and given by: \(a_n=2n\). Given \(n=4\) first positive even integers, then their sum equals to \(2+4+6+8=4(4+1)=20\).

For more check Number Theory chapter of Math Book: math-number-theory-88376.html
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Re: p/s sum of integers [#permalink]

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New post 09 Mar 2011, 17:22
Answer is C

Given sum of 1 2 3.....n = S

sum of 2 , 4 , 6 ....2n = 2[sum(1 2 3....n) ] = 2S

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Re: p/s sum of integers [#permalink]

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New post 12 Mar 2011, 14:18
hmm... lets say the series is

{1,2,3,4,5} then S = 15
the sum of the even numbers - 6.

what am i missing?
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Re: p/s sum of integers [#permalink]

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144144 wrote:
hmm... lets say the series is

{1,2,3,4,5} then S = 15
the sum of the even numbers - 6.

what am i missing?


The sum of the first 5 positive integers is 1+2+3+4+5=15;
The sum of the first 5 positive even integers is 2+4+6+8+10=30=2*15 (so the second sum shouldn't be the sum of the even numbers from the first list as you did).
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Re: p/s sum of integers [#permalink]

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New post 12 Mar 2011, 14:53
damn... got confused bc of the S.

thanks!
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Re: p/s sum of integers [#permalink]

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Lolaergasheva wrote:
If the sum of the first n positive integers is S, what is the sum of the first n positive even integers, in terms of S ?
(A) S/2
(B) S
(C) 2S
(D) 2S + 2
(E) 4S


I see that most people would solve it by taking some value of n... that's great... One suggestion - don't shy away from taking the lowest and simplest value.
I read 'sum of the first n positive integers', I say 'n = 1'... First 1 positive integer is 1 so S = 1. First 1 positive even integer will be 2 which is equal to twice of S. Only one options works out here so you are done. (In some cases, more than one option could work out e.g. if \(2S^2\) were another option. In that case you might want to check for n = 2 as well)
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Re: If the sum of the first n positive integers is S, what is [#permalink]

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New post 09 Jan 2017, 20:02
Here is what i did in this one =>
Let n=2 => S=1+2=3
Sum of first even numbers => 2+4=6
Clearly -> Sum = 2*S
Hence C.

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Re: If the sum of the first n positive integers is S, what is [#permalink]

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New post 08 Sep 2017, 20:52
Well, this one can be solved in less than 10 secs if we know some formulas

Sum of firs n natural numbers = n(n+1)/2

Sum of first n even natural numbers = n(n+1)

Sum of first n off natural numbers is n^2

ok
we are given that n(n+1)/2=S
that means n(n+1)=2S and we know that n(n+1) is sum of first n even natural numbers . And that's the answer.
2S
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Re: If the sum of the first n positive integers is S, what is   [#permalink] 08 Sep 2017, 20:52
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