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?

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.
