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

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

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

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  • $\begingroup$ 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 $\frac{i(i+1)}{2}$. $\endgroup$ – MathMajor Mar 27 '16 at 11:06
  • $\begingroup$ @BarryCipra no idea ^^ $\endgroup$ – tired Mar 27 '16 at 11:30
  • $\begingroup$ See also here. $\endgroup$ – Martin Sleziak Mar 29 '16 at 14:55
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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 $\sum_\limits{i=1}^n i$.

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  • $\begingroup$ a comment would be enough i think $\endgroup$ – tired Mar 27 '16 at 11:06
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    $\begingroup$ @tired Why? These two identities are well known and easily found online. As far as I am concerned, this is a complete answer. $\endgroup$ – MathMajor Mar 27 '16 at 11:08
  • $\begingroup$ @tired only if it helps the OP ... :-) $\endgroup$ – SchrodingersCat Mar 27 '16 at 11:08
  • $\begingroup$ @SchrodingersCat it is to trivial to really count for an answer ;) $\endgroup$ – tired Mar 27 '16 at 11:11
  • $\begingroup$ @tired The trivialness depends on how much and what the OP exactly knows...... ;) $\endgroup$ – SchrodingersCat Mar 27 '16 at 11:14
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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: $$\sum_{i=1}^n\frac{i(i+1)}2=\sum_{i=1}^{n}\binom{i+1}{2}=\sum_{i=2}^{n+1}\binom{i}{2}=\binom{n+2}{3}$$

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

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