1. The problem is to find the sum of the first $n$ terms of the series $1\cdot 2 + 2\cdot 3 + 3\cdot 4 + \cdots$. 2. Each term in the series can be written as $k(k+1)$ where $k$ is the term number starting from 1. 3. The sum of the first $n$ terms is therefore $S_n = \sum_{k=1}^n k(k+1)$. 4. Expand the term inside the summation: $k(k+1) = k^2 + k$. 5. So, $S_n = \sum_{k=1}^n (k^2 + k) = \sum_{k=1}^n k^2 + \sum_{k=1}^n k$. 6. Use the formulas for sums: - $\sum_{k=1}^n k = \frac{n(n+1)}{2}$ - $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$. 7. Substitute these into the sum: $$S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}.$$ 8. Factor out $\frac{n(n+1)}{6}$: $$S_n = \frac{n(n+1)}{6} (2n+1 + 3) = \frac{n(n+1)}{6} (2n+4).$$ 9. Simplify inside the parentheses: $$2n + 4 = 2(n+2).$$ 10. So, $$S_n = \frac{n(n+1)}{6} \times 2(n+2) = \frac{2n(n+1)(n+2)}{6} = \frac{n(n+1)(n+2)}{3}.$$ **Final answer:** $$S_n = \frac{n(n+1)(n+2)}{3}.$$