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Background

I wish to calculate $$ S= \sum_{i = 1}^{n}\frac{k(k+1)}{2}$$

I know what the answer is going to be, since this is essentially the sum of the first $n$ triangle numbers.

I.e. $S = (1) + (1+2) + (1+2+3) + \cdots + (1+2+3+\cdots+n)$

All solutions I've seen seem to know in advance what the answer is going to be, and their problem is proving it, which can be done using induction.

Question

However, I was wondering if this can be calculated using formulae for summations instead?

Alternatively, for instances where we do wish to prove it instead of calculate it, are there any other ways besides induction?

For reference

The answer should be $$\frac{n(n+1)(n+2)}{6}$$

Alec
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  • Relevant: http://math.stackexchange.com/questions/48080/prove-that-sum-limits-k-1nk2-fracnn12n16?noredirect=1&lq=1 (many non-inductive proofs) – Clement C. Aug 28 '16 at 12:43
  • @ClementC.- That's roughly what I'm looking for, yes. I guess a lot of the same methods can be applied to this one? – Alec Aug 28 '16 at 12:47
  • Since what you want to compute is $\frac{1}{2}\sum_{i=1}^n k^2+\frac{1}{2}\sum_{i=1}^n k$, yes... this tells you how to compute the first term. – Clement C. Aug 28 '16 at 12:48

6 Answers6

7

Instead of working forwards, work backwards:

$$\begin{align} \sum_{k=1}^n\frac{k(k+1)}{2} & = \sum_{k_1=1}^n\sum_{k_2=1}^{k_1}k_2 \\ & = \sum_{k_1=1}^n\sum_{k_2=1}^{k_1}\sum_{k_3=1}^{k_2}1 \\ \end{align}$$

This can be thought as a combinatoric problem and simply reduces to:

$$\frac{(n+2)!}{3!(n-1)!}=\frac{n(n+1)(n+2)}{6}$$

It is the number of ways you can choose $3$ elements from $n+2$ elements. For example, the number of ways you can choose $3$ letters from $\lbrace a,b,c,d,e,f\rbrace$ can be shown as

$$\underbrace{(abc,abd,abe,abf)}_4,\underbrace{(acd,ace,acf)}_3,\underbrace{(ade,adf)}_2,\underbrace{(aef)}_1$$

The number of ways can be thought of as $4+3+2+1=1+(1+2)+(1+2+3)=10$ or $_5C_3=\frac{5!}{3!(5-3)!}=\frac {120}{12}=10$

This method can be generalized to

$$\sum_{k_1=1}^r\sum_{k_2=1}^{k_1}\sum_{k_3=1}^{k_2}\dots\sum_{k_{n-1}=1}^{k_{n-2}}\sum_{k_n=1}^{k_{n-1}}1=\frac{(r+n-1)!}{n!(r-1)!}$$

which means the number of ways to choose $n$ elements from $r+n$ elements.

Simply Beautiful Art
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    elegant, loved this answer ;) – H G Sur Aug 28 '16 at 12:49
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    @HGSur tips the hat – Simply Beautiful Art Aug 28 '16 at 12:50
  • I don't see why $$\sum_{k_1=1}^n\sum_{k_2=1}^{k_2}\sum_{k_3=1}^{k_2} 1=\frac{n!}{3!(n-3)!}.$$ – Surb Aug 28 '16 at 12:52
  • @Surb Its the number of ways you can pick $3$ different unique elements from $n$ elements. – Simply Beautiful Art Aug 28 '16 at 12:55
  • I trust you, but I don't see why. Do you have any reference ? – Surb Aug 28 '16 at 12:56
  • @Surb Try counting the number of ways you can pick $3$ letters from $\lbrace a,b,c,d,ef\rbrace$. Counting like this, you get: $$(abc,abd,abe,abf),(acd,ace,acf),(ade,adf),(aef)$$The grouping of represents the sum of the sum of the sum of $1$, but it is also known to be more simply $\frac{5!}{3!(5-3)!}=10$ different ways, which is agreement if you count the result above. Similarly, looking at it backwards, this is the same as $1+2+3+4$, which is what was asked for. I'm still looking for a reference if you would still like. – Simply Beautiful Art Aug 28 '16 at 13:11
  • If you can find a reference, I would appreciate. But very good answer anyways :-) (+1) – Surb Aug 28 '16 at 13:23
  • @Surb I'm almost sorry because I feel like I can't find a reference even though it seems like this is what I would call the counting definition of a combination. No one states the formula explicitly >:( quite frustrating – Simply Beautiful Art Aug 28 '16 at 13:26
  • @HGSur You got a formal explanation? – Simply Beautiful Art Aug 28 '16 at 13:32
  • @SimpleArt Sorry, I can't find any either – H G Sur Aug 28 '16 at 13:47
4

If you know: $$ \sum\limits_{k=1}^n k = \frac{n(n+1)}{2} $$ and $$ \sum\limits_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} $$ then: $$ \sum\limits_{k=1}^n \frac{k(k+1)}{2} = \frac{1}{2} \left[\sum\limits_{k=1}^n k^2 + \sum\limits_{k=1}^n k \right] = \frac{1}{2}\left[\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}\right] = \frac{n(n+1)(n+2)}{6} $$

Tom
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  • This is very close to what I was attempting, but I think I blundered by assuming that the summation was distributive over a product. – Alec Aug 28 '16 at 13:04
  • @Alec Did this clear up your confusion then? Just to note, I made an edit since I erroneously wrote $k$ where I should have written $n$. – Tom Aug 28 '16 at 13:07
  • Yes, this cleared it up perfectly. My blunder was trying to equate $$\sum\limits_{k=1}^n \frac{k(k+1)}{2} = \frac12 \sum\limits_{k=1}^n k \sum\limits_{k=1}^n(k+1)$$ when in fact I should have expanded the parenthesis instead. – Alec Aug 28 '16 at 15:22
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A telescopic approach: $$S= \sum_{i = 1}^{n}\frac{k(k+1)}{2}= \frac{1}{6}\sum_{i = 1}^{n}[(k+1)^3-k^3-1]= \frac{1}{6}\left[\sum_{i = 1}^{n}[(k+1)^3-k^3]-n\right]\\=\frac{1}{6}\left[(n+1)^3-1-n\right]=\frac{(n+1)}{6}\left[(n+1)^2-1\right]=\frac{(n+1)(n+2)n}{6}.$$

Robert Z
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Hint

$$\frac{k(k+1)}{2}=\sum_{j=1}^k j.$$ Therefore, your sum is $$\sum_{k=1}^n\sum_{j=1}^k j=\sum_{j=1}^n\sum_{k=j}^n j=\sum_{j=1}^n (n-j+1)j=n\sum_{j=1}^nj-\sum_{j=1}^n j^2+\sum_{j=1}^n j.$$

You have all elements to answer.

Surb
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Another way to deal with your sum, and in general with the sum of the binomial coefficient in the upper index, is to consider that $$ \Delta _{\,n} \left( \begin{gathered} n \\ m \\ \end{gathered} \right) = \left( \begin{gathered} n + 1 \\ m \\ \end{gathered} \right) - \left( \begin{gathered} n \\ m \\ \end{gathered} \right) = \left( \begin{gathered} n \\ m - 1 \\ \end{gathered} \right) $$ So $$ \sum\limits_{a\,\, \leqslant \,k\, \leqslant \,b} {\left( \begin{gathered} k \\ m \\ \end{gathered} \right)} = \sum\limits_{a\,\, \leqslant \,k\, \leqslant \,b} {\Delta _{\,n} \left( \begin{gathered} k \\ m + 1 \\ \end{gathered} \right)} = \left( \begin{gathered} b + 1 \\ m + 1 \\ \end{gathered} \right) - \left( \begin{gathered} a \\ m + 1 \\ \end{gathered} \right) $$ and in your case $$ \sum\limits_{1\,\, \leqslant \,k\, \leqslant \,n} {\left( \begin{gathered} k + 1 \\ 2 \\ \end{gathered} \right)} = \sum\limits_{2\,\, \leqslant \,j\, \leqslant \,n + 1} {\left( \begin{gathered} j \\ 2 \\ \end{gathered} \right)} = \left( \begin{gathered} n + 2 \\ 3 \\ \end{gathered} \right) - \left( \begin{gathered} 2 \\ 3 \\ \end{gathered} \right) = \left( \begin{gathered} n + 2 \\ 3 \\ \end{gathered} \right) $$

G Cab
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You are summing Binomial coefficients. In this case the relevant identity is $$\binom{2}{2}+\binom{3}{2}+\cdots +\binom{n}{2}=\binom{n+1}{3}$$ This method admits much generalization.

For example $n^3=\binom{n}{1}+6\binom{n}{2}+6\binom{n}{3}$

Rene Schipperus
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