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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Answer is C

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

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

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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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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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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