Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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
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 $n$. $\square$

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.