1. **State the problem:** We need to 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, where $n$ is an integer. 2. **Write the sum:** $$\frac{1}{2}n(n+1) + \frac{1}{2}(n+1)(n+2)$$ 3. **Factor out the common term $\frac{1}{2}(n+1)$:** $$\frac{1}{2}(n+1)(n + n + 2) = \frac{1}{2}(n+1)(2n + 2)$$ 4. **Simplify inside the parentheses:** $$\frac{1}{2}(n+1) \cdot 2(n+1)$$ 5. **Cancel the 2 in numerator and denominator:** $$(n+1)(n+1) = (n+1)^2$$ 6. **Conclusion:** The sum simplifies to $$(n+1)^2$$ which is a perfect square for any integer $n$. Therefore, the sum is always a square number.