Multiply out (k)(k+ 3) and use the formulas for the sum of the first n natural numbers and for the first n square numbers. Alternatively, use mathematical induction. I can write both out if it would be helpful.

$\sum_{k=1}^{n} k(k+3) = \sum_{k=1}^{n} (k^{2} + 3k)$ $= \sum_{k=1}^{n} k^{2} + 3\sum_{k=1}^{n} k$ $= \frac {k(k+1)(2k+1)}{6} + \frac{3k(k+1)}{2}$ $=\frac{k(k+1)(k+5)}{3}$

Base case: $1 \cdot 4 = \frac{1 \cdot 2 \cdot 6}{3} =4$ Assume true for $n = k$, then adding $(n+1)(n+4)$ to both sides, we get $\frac{n(n+1)(n+5)}{3} + \frac{3(n+1)(n+4)}{3}$ $=\frac{(n+1)(n^{2} + 8n + 12)}{3}$ $= \frac{(n+1)(n+2)(n+6)}{3}$ $=\frac{(n+1)(n+1+1)(n+1+5)}{3}$