Double summation counting I have the following:
$$T_n=\sum_{j=1}^n\left(\sum_{k=1}^jk\right)$$
I know that the counting of numbers $1$ to $n$ can be expressed as $$\frac{(n+1)n}{2},$$ which leaves me with $$T_n=\sum_{j=0}^n\left(\frac{(j+1)j}{2}\right).$$
I am trying to simplify $T_n$ as to write it without summations, and I am not entirely sure if this is possible. I've found that, if $T_n$ is in standard form in the numerator ($\frac{aj^2+bj+c}{2}$), the following is true:
$$\begin{align}a&=n\\b&=\sum_{q=0}^{n-1}2q+1\end{align}$$
I have been thus far unable to find a pattern for $c$. Can any of you help me? Thanks!
 A: For sure, $T_n$ is a third-degree polynomial in $n$, hence you may find its coefficients by interpolation, once computed $T_0,T_1,T_2,T_3$. Or you may notice that:
$$ T_n = \sum_{j=1}^{n}\sum_{1\leq k\leq j}k =\sum_{j=1}^{n}\binom{j+1}{2}=\binom{n+2}{3}=\color{red}{\frac{n(n+1)(n+2)}{6}}\tag{1}$$
by a well-known combinatorial identity. 
You may also notice that $T_n$ is the number of lattice points in the region:
$$ \{(x,y,z)\in \mathbb{N}^3 : x+y+z\leq n-1\}.\tag{2}$$
A: Sure you can simplify it: 
$$T_n=\sum_{j=0}^n\left(\frac{(j+1)j}{2}\right)= \frac{1}{2}\sum_{j=0}^n\left((j^2+j)\right)= \frac{1}{2}\sum_{j=0}^nj+ \frac{1}{2}\sum_{j=0}^nj^2$$ 
I think you are able to go from here
A: Hint. You may find Faulhaber's formulas useful. In particular, we have
$$
\sum_{k=1}^n k^2=\frac{n(n+1)(2\,n+1)}{6}
$$
A: Other way!
Changer id this sum
$$T_n=\sum_{j=1}^n \sum_{k=1}^j k=\sum_{k=1}^n\sum_{j=k}^n k=\sum_{k=1}^n(k(n+1-k))\quad\textit{because $(j\ge k)$}$$
$$\Rightarrow T_n=(n+1)\sum_{k=1}^n k -\sum_{k=1}^n k^2$$
More
$$2T_n=\sum_{j=1}^n(j+j^2)=\sum_{k=1}^n k +\sum_{k=1}^n k^2$$
Addition $\quad\displaystyle 3T_n=(n+2)\sum_{k=1}^n k \Rightarrow T_n=\frac{(n+2)n(n+1)}{6}$
