1. **State the problem:** Prove that for every positive integer $n$, the sum $$\sum_{k=1}^n \frac{1}{(k+2)(k+3)} = \frac{n}{3n+9}.$$ 2. **Rewrite the general term:** Notice that $$\frac{1}{(k+2)(k+3)} = \frac{1}{k+2} - \frac{1}{k+3}$$ by partial fraction decomposition: \[ \frac{1}{(k+2)(k+3)} = \frac{A}{k+2} + \frac{B}{k+3}. \] Setting up the equation: $$1 = A(k+3) + B(k+2).$$ Let $k = -2$, then $1 = A(1) + B(0) \implies A = 1$. Let $k = -3$, then $1 = A(0) + B(-1) \implies B = -1$. So partial fractions give: $$\frac{1}{(k+2)(k+3)} = \frac{1}{k+2} - \frac{1}{k+3}.$$ 3. **Sum the series:** Substitute into the sum: $$\sum_{k=1}^n \frac{1}{(k+2)(k+3)} = \sum_{k=1}^n \left( \frac{1}{k+2} - \frac{1}{k+3} \right).$$ 4. **Telescoping cancellation:** Writing out the first few terms: $$\left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \dots + \left( \frac{1}{n+2} - \frac{1}{n+3} \right).$$ Most terms cancel out, leaving: $$\frac{1}{3} - \frac{1}{n+3}.$$ 5. **Simplify the remaining expression:** $$\frac{1}{3} - \frac{1}{n+3} = \frac{n+3 - 3}{3(n+3)} = \frac{n}{3(n+3)}.$$ 6. **Rewrite denominator:** Notice that $3(n+3) = 3n + 9$, so $$\sum_{k=1}^n \frac{1}{(k+2)(k+3)} = \frac{n}{3n + 9},$$ which completes the proof.