# Summing $r(r+3)$ using induction

We want to prove the following summation by induction: $$\sum_{r=1}^{n}r(r+3)=\frac{1}{3}n(n+1)(n+5)$$ The problem is posted for a friend, but others can look at the solution if they want/need.

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Let us denote the sum by $S_{n}$. We first establish the base case, that $S_{1}=1(1+3)=4$ is equal to $\frac{1}{3}\cdot 1\cdot 2\cdot 6=4$, which holds.

Then, assume that for some $k$, that $S_{k}=\frac{1}{3}k(k+1)(k+5)$. Obviously $S_{k+1}=S_{k}+(k+1)(k+4).$ Tidying up, we have
$$S_{k+1}=\frac{1}{3}(k+1)\left[k^{2}+5k+3k+12\right]=\frac{1}{3}(k+1)(k^{2}+8k+12)=\frac{1}{3}(k+1)(k+2)(k+6)$$
Which is as we expected. If $S_{k}$ is given by the formula above, then $S_{k+1}$ is. Therefore, since the claim holds for $n=1$, it holds for all natural $n$. $\square$

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Just an addition to Daniel's answer. There is a slightly different way of proving such values, called perturbation method (see Concrete Mathematics, Chapter 2 by Graham, Knuth and Patashnik). Clearly the value you want is

$V_n = \sum_{k=1}^{n}k(k+3)=\sum_{k=1}^{n}k^2 + 3 \sum_{k=1}^{n}k=S^{2}_{n}+3 S^{1}_{n}$, here 2 and 1 are superscripts for convenience.

Now look at $S^{2}_n = \sum_{k=1}^{n}k^2$, hence $S^{2}_{n+1}=S^{2}_{n} +(n+1)^2=\sum_{k=1}^{n}(k+1)^2+1=S^{2}_{n}+2 \sum_{k=1}^{n}+n+1$. From this we can easily find by rearranging terms that $$S^{1}=\frac{n(n+1)}{2}$$ This solves the second problem. Now let's look at $$S^{3}_{n}=\sum_{k=1}^{n}k^3 \Leftrightarrow S^{3}_{n}+(n+1)^3=\sum_{k=1}^{n}(k+1)^3+1=S^{3}_{n}+3S^{2}_{n}+3S^{1}_{n}+1$$ Once again, after some algebra we find that $$S^{2}_n=\frac{(n+1)^3}{3}-\frac{n(n+1)}{2}- \frac{n+1}{3}$$ Now we have both results for $V_n$. Plug them in the expression above, do some algebra adn you will get the desired result.

I should say this method is quite universal.

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