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As we know, triangular numbers are a sequence defined by $\frac{n(n+1)}{2}$. And it's first few terms are $1,3,6,10,15...$. Now I want to calculate the sum of the sum of triangular numbers. Let's define $$a_n=\frac{n(n+1)}{2}$$ $$b_n=\sum_{x=1}^na_x$$ $$c_n=\sum_{x=1}^nb_x$$ And I want an explicit formula for $c_n$. After some research, I found the explicit formula for $b_n=\frac{n(n+1)(n+2)}{6}$. Seeing the patterns from $a_n$ and $b_n$, I figured the explicit formula for $c_n$ would be $\frac{n(n+1)(n+2)(n+3)}{24}$ or $\frac{n(n+1)(n+2)(n+3)}{12}$.

Then I tried to plug in those two potential equations,

If $n=1$, $c_n=1$, $\frac{n(n+1)(n+2)(n+3)}{24}=1$, $\frac{n(n+1)(n+2)(n+3)}{12}=2$. Thus we can know for sure that the second equation is wrong.

If $n=2$, $c_n=1+4=5$, $\frac{n(n+1)(n+2)(n+3)}{24}=5$. Seems correct so far.

If $n=3$, $c_n=1+4+10=15$, $\frac{n(n+1)(n+2)(n+3)}{24}=\frac{360}{24}=15$.

Overall, from the terms that I tried, the formula above seems to have worked. However, I cannot prove, or explain, why that is. Can someone prove (or disprove) my result above?

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    $\begingroup$ They're diagonals in Pascal's Triangle. $\endgroup$ Jun 25 '21 at 15:05
  • $\begingroup$ @JMoravitz I think that's way off. I am dealing with triangular numbers not square numbers here. Also my question is actually a double sum not a single one. $\endgroup$ Jun 25 '21 at 15:12
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    $\begingroup$ @JMoravitz There is a more direct answer. $\endgroup$
    – Jean Marie
    Jun 25 '21 at 15:14
  • $\begingroup$ @JeanMarie I saw this post before. That's how I got $\frac{n(n+1)(n+2)}{6}$. However, I want the sum of this sequence. $\endgroup$ Jun 25 '21 at 15:15
  • $\begingroup$ In general, $\prod\limits_{k=0}^{p-1} (n+k) = \frac{(n+p) - (n-1)}{p+1}\prod\limits_{k=0}^{p-1} (n+k) = \frac{1}{p+1}\left(\prod\limits_{k=0}^p(n+k) - \prod\limits_{k=0}^p(n-1 + k)\right)$. Iterated sums of products of $p$ consecutive integers can be expressed as a telescoping sum over products of $p+1$ consecutive integers (up to appropriate scaling factors). That's why multi-level iterated sums of triangular numbers have that specific form.... $\endgroup$ Jun 25 '21 at 15:24
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The easiest way to prove your conjecture is by induction. You already checked the case $n=1$, so I won’t do it again. Let’s assume your result is true for some $n$. Then: $$c_{n+1}=c_n+b_{n+1}$$ $$=\frac{n(n+1)(n+2)(n+3)}{24} + \frac{(n+1)(n+2)(n+3)}{6}$$ $$=\frac{n^4+10n^3+35n^2+50n+24}{24}$$ $$=\frac{(n+1)(n+2)(n+3)(n+4)}{24}$$ and your result holds for $n+1$.

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This can be generalized, in fact if $U_p(n)=(n+1)(n+2)\cdots(n+p)$ then we have the summation formula (proved here)

$$\sum\limits_{k=1}^n U_p(k)=\frac{1}{p+2}\,U_{p+1}(n)$$

In particular, it is a bit of a pity to see answers in which $\sum i$, $\sum i^2$ and $\sum i^3$ are separated, because this is kind of going against the natural way of solving it.

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Hint: $$\sum_{r=1}^n r=\frac{n(n+1)}{2}$$ $$\sum_{r=1}^n r^2=\frac {n(n+1)(2n+1)}{6}$$ $$\sum_{r=1}^n r^3=\frac {(n(n+1))^2}{4}$$ Use of these $3$ formulae is sufficient to prove the required result.

The derivation of the $3^{rd}$ formula can comes by noting: $$(r+1)^4-r^4=4r^3+6r^2+4r+1$$ Now sum this identity over $r=1$ to $r=n$, and since $\sum r^2$ and $\sum r$ are already known, the $3^{rd}$ formula gets proven. In general, using this process, $\sum r^n$ can be derived if $\sum r^{n-1}$ is known.

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  • $\begingroup$ Wait so is $$\sum_{r=1}^{n}r^3=\left(\sum_{r=1}^{n}r\right)^2$$ $\endgroup$ Jun 25 '21 at 15:04
  • $\begingroup$ Yes, indeed. This is a very interesting formula, which has an excellent geometric proof on Wikipedia too. $\endgroup$ Jun 25 '21 at 15:06
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Notice that after $k$ summations, the formula is

$$\binom{n+k-1}{n-1}.$$

As we can check, by the Pascal identity

$$\binom{n+k-1}{n-1}-\binom{n-1+k-1}{n-2}=\binom{n-1+k-1}{n-1},$$

which shows that the last term of a sum (sum up to $n$ minus sum up to $n-1$) is the sum of the previous stage ($k-1$) up to $n$.

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Have you tried using induction to prove or disprove your attempts? It tends to be relevant with these equations.

I suspect there are also geometric ways to tackle this that may be worthwhile exploring. Roger Fenn's Geometry has some problems of this nature.

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One approach is to calculate $5$ terms of $c_n$, recognize that it's going to be a degree-4 formula, and then solve for the coefficients. Thus:

$$c_1 = T_1=1 \\ c_2 = c_1 + (T_1+T_2) = 5 \\ c_3 = c_2+(T_1+T_2+T_3) = 15 \\ c_4 = c_3 + (T_1+T_2+T_3+T_4) = 35 \\ c_5 = c_4 + (T_1+T_2+T_3+T_4+T_5) = 70$$ Now we can find coefficients $A,B,C,D,E$ so that $An^4+Bn^3+Cn^2+Dn+E$ gives us those results when $n=1,2,3,4,5$. This leads to a linear system in 5 unknowns, which we can solve and obtain $A=\frac1{24},B=\frac14,C=\frac{11}{24},D=\frac14,E=0$. Thus taking a common denominator, we have $$c_n=\frac{n^4+6n^3+11n^2+6n}{24}=\frac{n(n+1)(n+2)(n+3)}{24}$$ So that agrees with your result.


Another way is to use the famous formulas for sums of powers. Thus, we find $b_n$ first: $$b_n = \sum_{i=1}^n \frac{i(i+1)}{2} = \frac12\left(\sum i^2 + \sum i\right) = \frac12\left(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right)\\ =\frac{n^3+3n^2+2n}{6}$$

Now, we find $c_n$: $$c_n = \sum_{i=1}^n \frac{i^3+3i^2+2i}{6}=\frac16\sum i^3 + \frac12\sum i^2 + \frac13\sum i \\ = \frac16\frac{n^2(n+1)^2}{4} + \frac12\frac{n(n+1)(2n+1)}{6} + \frac13\frac{n(n+1)}{2} \\ = \frac{n^4+6n^3+11n^2+6n}{24}=\frac{n(n+1)(n+2)(n+3)}{24}$$

So we have confirmed the answer 2 different ways. As is clear from the other solutions given here, there are other ways as well.

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