Let's denote the smallest even number in the series of three consecutive even numbers as \( n \). Since these are even numbers and they are consecutive, it means that they increase by 2 as we go from one to the next. Therefore, the middle even number can be represented as \( n + 2 \), and the largest even number as \( n + 4 \).
The problem statement asks us to prove that the sum of the squares of the smallest and the largest numbers is 8 more than twice the square of the middle number.
Let us express algebraically what we need to prove:
\[
n^2 + (n + 4)^2 \stackrel{?}{=} 2 \times (n + 2)^2 + 8
\]
Now we will work on both sides of this equation separately to try and prove if they are indeed equal.
1. *Calculate the sum of the squares of the smallest and the largest numbers*:
\[
\begin{align*}
n^2 + (n + 4)^2 &= n^2 + (n^2 + 8n + 16)\\
&= n^2 + n^2 + 8n + 16\\
&= 2n^2 + 8n + 16
\end{align*}
\]
2. *Calculate twice the square of the middle number plus 8*:
\[
\begin{align*}
2 \times (n + 2)^2 + 8 &= 2 \times (n^2 + 4n + 4) + 8\\
&= 2n^2 + 8n + 8 + 8\\
&= 2n^2 + 8n + 16
\end{align*}
\]
By comparing the two results:
\[
2n^2 + 8n + 16 = 2n^2 + 8n + 16
\]
we see that they are indeed the same. This algebraically confirms that, for any three consecutive even numbers, the sum of the squares of the smallest and the largest is exactly 8 more than twice the square of the middle number.
Hence, the initial statement is proved.
