If the sum of the even integers between 1 and k inclusive is equal to 2k, what is the value of k?

How do you prove that k must be 6 and not larger or smaller than 6?

inclusive? 1 is not an even integer.

note the sum of the first "$n$" even integers is ...

$2+4+6+...+2n=n(n+1)$

in this case, $k = 2n \implies n = \frac{k}{2}$, so the sum of the even integers between 2 and k inclusive is $\displaystyle \frac{k}{2}\left(\frac{k}{2}+1\right) = 2k$

solving for $k$ ...

\(\displaystyle \frac{k^2}{4} + \frac{k}{2} = 2k\)

\(\displaystyle k^2 + 2k = 8k\)

\(\displaystyle k^2 - 6k = 0\)

\(\displaystyle k(k - 6) = 0\)

$k = 0$ (which can't be because $k > 2$ ...)

\(\displaystyle k = 6\)