Find the sum of the series below Find the sum

$$(1\cdot2)+(1\cdot3)+(1\cdot4)+\cdots+(1\cdot2015)+(2\cdot3)+(2\cdot4)+\cdots+(2\cdot2015)+\cdots+(2014\cdot2015)$$

What I have tried...
We are looking for $$S(n)=\sum_{i=1}^{n-1}\sum_{j=i+1}^nij$$ when $n=2015$
$$S(n)=\sum_{1\le i<j\le n}ij$$ $$=\frac 12\left((\sum_{i=1}^n i)^2-\sum_{i=1}^ni^2\right)$$
$$S(n)=\frac 12\left(\frac{n^2(n+1)^2}4-\frac{n(n+1)(2n+1)}6\right)$$
 $$=\frac{(n-1)n(n+1)(3n+2)}{24}$$
And so $$\boxed{S(2015)=\frac{2014\times 2015\times 2016\times 6047}{24}}$$
Is the solution wrong or not?
 A: Summing along diagonals gives a nice symmetric derivation of the solution, using binomial coefficients and minimizing messy manipulation of fractions.
$$\begin{align}
S(n)&=\sum_{i=1}^{n-1}\sum_{j=i+1}^nij
\color{lightgray}{=\sum_{1\le i<j\le n} ij}\\\\
&=\begin{cases}\begin{matrix}
\color{orange}{1\cdot 2} &+1\cdot 3 &+1\cdot 4 &+\cdots &+\color{purple}{1\cdot (n-2)} &\color{blue}{+1\cdot (n-1)} &\color{green}{+1\cdot n}&\\
          &\color{orange}{+2\cdot 3}  &+2\cdot 4 &+\cdots &+2\cdot (n-2) &\color{purple}{+2\cdot (n-1)} &\color{blue}{+2\cdot n}&\\
          &           &\color{orange}{+3\cdot 4} &+\cdots &+3\cdot (n-2) &+3\cdot (n-1) &\color{purple}{+3\cdot n}& \\
          &           &          &\ddots  &              &            &\vdots  &\\
          &           &          &        &              &             &\color{orange}{+(n-1)\cdot n}&
\end{matrix}\end{cases}\\\\
&\color{lightgray}{=\sum_{r=1}^{n-1}\sum_{i=1}^{n-r}i(i+r)
\qquad\qquad\text{using}\ r=j-i}\\
&\color{lightgray}{=\sum_{m=1}^{n-1}\sum_{i=1}^mi(n-m+i)
\qquad\text{using}\ m=n-r}\\\\
&=\begin{cases}\begin{array}{r}
\color{green}{1\cdot n}&\\
\color{blue}{+1\cdot (n-1)+2\cdot n}&\\
\color{purple}{+1\cdot (n-2)+2\cdot (n-1)+3\cdot n}&\\
\vdots\\
\color{orange}{+1\cdot 2+2\cdot3 +3\cdot 4+\cdots +(n-3)\cdot (n-2)+(n-2)\cdot (n-1)+(n-1)\cdot n}\end{array}\end{cases}\\\\
&\color{lightgray}{=\sum_{m=1}^{n-1}\sum_{i=1}^m ni-i(m-i)}\\
&\color{lightgray}{=\sum_{m=1}^{n-1}\left[\binom{m+1}2 n-\sum_{m=1}^m\sum_{k=i+1}^m i\right]}\\
&\color{lightgray}{=\sum_{m=1}^{n-1}\left[\binom{m+1}2 n-\sum_{k=2}^m\sum_{i=1}^{k-1}i\right]}\\
&\color{lightgray}{=\sum_{m=1}^{n-1}\left[\binom{m+1}2 n-\sum_{k=2}^m\binom k2\right]}\\
&\color{lightgray}{=\sum_{m=1}^{n-1}\left[\binom{m+1}2 n-\binom {m+1}3\right]}\\
&=\begin{cases}\begin{array}{l}\quad 
\color{Green}{n}&\\
\color{blue}{+3n-1}\\
\color{purple}{+6n-4}&\\
+10n-10&\\
\vdots\\
+\binom {m+1}2n-\binom {m+1}3&\\
\vdots \\
\color{orange}{+\binom n2 n-\binom n3}\end{array}\end{cases}\\\\
&=\sum_{m=1}^{n-1}\binom {m+1}2 n-\binom {m+1}3\\\\
&=\binom {n+1}3n-\binom {n+1}4\\\\
&=n\binom {n+1}3-\frac{n-2}4\binom{n+1}3\\\\
&=\frac{3n+2}4\binom{n+1}3\qquad\blacksquare\\\\
S(2015)&=\frac {6047}4\binom{2016}3=2\;061\;359\;653\;080\qquad\blacksquare
\end{align}$$

Note
The summation can also be computed by summing columns instead of rows (i.e. swapping $i$ and $j$) as follows:
$$\begin{align}
S(n)
&=\sum_{i=1}^{n-1}\sum_{j=i+1}^n ij\\
&=\sum_{1\le i<j\le n}ij\\
&=\sum_{j=2}^n\sum_{i=1}^{j-1}ij\\
&=\sum_{j=1}^n j\sum_{i=1}^{j-1}\binom i1\\
&=\sum_{j=1}^n j\binom j2\\
&=\sum_{j=1}^n (j+1)\binom j2-\binom j2\\
&=\sum_{j=1}^n 3\binom {j+1}3-\binom j2\\
&=3\binom {n+2}4-\binom {n+1}3\\
&=\frac {3(n+2)}4\binom {n+1}3-\binom {n+1}3\\
&=\frac {3n+2}4\binom {n+1}3\qquad\blacksquare
\end{align}$$
