# sum of series $\sum \limits_{i=1}^{n}\frac{i(i+1)}{2}$ [closed]

Does there exist an explicit formula for the sum of the series

$$\sum \limits_{i=1}^{n}\frac{i(i+1)}{2}$$

## closed as off-topic by Did, tired, Shailesh, choco_addicted, Daniel W. FarlowMar 27 '16 at 11:43

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• Kind of a funny question for some reason ... I was about to edit an equals sign in to your post until I realized that we were actually summing $\frac{i(i+1)}{2}$. – MathMajor Mar 27 '16 at 11:06
• @BarryCipra no idea ^^ – tired Mar 27 '16 at 11:30
• See also here. – Martin Sleziak Mar 29 '16 at 14:55

## 2 Answers

HINT:

A good way is to simplify the summation as follows: $$\sum_{i=1}^n \frac{i(i+1)}{2}=\frac{1}{2}\left[\sum_{i=1}^n i^2 + \sum_{i=1}^n i \right]$$ and then use the standard results for $\sum_\limits{i=1}^n i^2$ and $\sum_\limits{i=1}^n i$.

• a comment would be enough i think – tired Mar 27 '16 at 11:06
• @tired Why? These two identities are well known and easily found online. As far as I am concerned, this is a complete answer. – MathMajor Mar 27 '16 at 11:08
• @tired only if it helps the OP ... :-) – SchrodingersCat Mar 27 '16 at 11:08
• @SchrodingersCat it is to trivial to really count for an answer ;) – tired Mar 27 '16 at 11:11
• @tired The trivialness depends on how much and what the OP exactly knows...... ;) – SchrodingersCat Mar 27 '16 at 11:14

In general we have:$$\sum_{i=r}^{n}\binom{i}{r}=\binom{n+1}{r+1}$$ This is a useful equality that can be proved with induction on $n$.

The induction step is (triangle of Pascal): $$\binom{n+1}{r+1}+\binom{n+1}{r}=\binom{n+2}{r+1}$$

You can apply this to find: $$\sum_{i=1}^n\frac{i(i+1)}2=\sum_{i=1}^{n}\binom{i+1}{2}=\sum_{i=2}^{n+1}\binom{i}{2}=\binom{n+2}{3}$$

• Does the edit please you?.. :) – SchrodingersCat Mar 27 '16 at 11:25
• Very nice identities, yet if the asker has the level his question implies, I think most probably doesn't know this. +1 – DonAntonio Mar 27 '16 at 11:27
• @Joanpemo Thanks. Well, let's hope he (and others) will learn something of this. – drhab Mar 27 '16 at 11:30
• Wow. I didn't get the idea of approaching this through combinations at all! I just did the normal brute force expansion. – user230452 Mar 27 '16 at 11:38
• @SchrodingersCat Sure. It certainly is a good way. – drhab Mar 27 '16 at 11:45