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

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