<s>[INST] How do I calculate the sum of sum of triangular numbers? 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?
 [/INST] 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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