\section*{Problem 4 - 051044}

Calculate the difference \(D\) between the sum of the squares of all even natural numbers \(\leq 100\) and the sum of the squares of all odd natural numbers \(<100\)!
}

It holds: Sum of the squares of odd numbers \(<100=\sum_{i=1}^{50}(2 i-1)^{2}\) and sum of the squares of even numbers \(\leq 100=\sum_{i=1}^{50}(2 i)^{2}\). Therefore, it follows that

\[
\begin{gathered}
D=\sum_{i=1}^{50}(2 i)^{2}-\sum_{i=1}^{50}(2 i-1)^{2}=\sum_{i=1}^{50} 4 i^{2}-\sum_{i=1}^{50}\left(4 i^{2}-4 i+1\right)=-50+\sum_{i=1}^{50} 4 i= \\
=-50+4 \cdot \frac{50}{2} \cdot 51=5050
\end{gathered}
\]