\( n \) is an integer. Prove algebraically that the sum of \( \frac{1}{2} n(n+1) \) and \( \frac{1}{2}(n+1)(n+2) \) is always a square number.

To prove that the sum of \( \frac{1}{2} n(n+1) \) and \( \frac{1}{2}(n+1)(n+2) \) is always a square number, we start by calculating the sum. First, we rewrite each term: \[ \frac{1}{2} n(n+1) + \frac{1}{2} (n+1)(n+2) \] Factoring out \( \frac{1}{2} \), we get: \[ \frac{1}{2} \left( n(n+1) + (n+1)(n+2) \right) \] Next, we simplify \( n(n+1) + (n+1)(n+2) \): \[ n(n+1) = n^2 + n \] And for \( (n+1)(n+2) \): \[ (n+1)(n+2) = n^2 + 3n + 2 \] Adding these, we have: \[ n(n+1) + (n+1)(n+2) = (n^2 + n) + (n^2 + 3n + 2) = 2n^2 + 4n + 2 \] Now, putting it back into our factored equation: \[ \frac{1}{2} (2n^2 + 4n + 2) = n^2 + 2n + 1 \] Notice that: \[ n^2 + 2n + 1 = (n+1)^2 \] Thus, we conclude that the sum \[ \frac{1}{2} n(n+1) + \frac{1}{2}(n+1)(n+2) = (n+1)^2 \] is always a perfect square, specifically the square of \( n+1 \). Therefore, we have successfully proven that this sum is always a square number: \[ \boxed{(n+1)^2} \]