Multiply And Add

$\Large \displaystyle\sum_{n=1}^{2015}(n-1)(n+1)$ $\Large =\displaystyle\sum_{n=1}^{2015}(n^2-1)$ $\Large =\left(\displaystyle\sum_{n=1}^{2015}n^2\right)-2015$ $\Large=\left(\dfrac{(2015)(2016)(4031)}{6}\right)-2015$ $\huge =\boxed{\color{#007fff}{2729146225}}$ $\color{#D61F06}{\boxed{\mathcal{\color{#0C6AC7}{\text{Formula Used:}}}\\ \color{forestgreen}{\displaystyle\sum n^2=\dfrac{n(n+1)(2n+1)}{6}}}}$

We can rewrite this expression as:

$\begin{aligned} \color{#007fff}{\huge\sum_{n=1}^{2014} n(n+2)}&\huge=\sum_{n=1}^{2014} \color{#D61F06}{n^2}+\color{#20A900}{2n}\\ \huge&\huge=\sum_{n=1}^{2014} \color{#D61F06}{n^2} +\sum_{n=1}^{2014} \color{#20A900}{2n}\\ \huge&\huge=\color{#D61F06}{\dfrac{n(n+1)(2n+1)}6}+\color{#20A900}{2\times \dfrac{n(n+1)}2}\\ \huge&\huge=\color{#D61F06}{\dfrac{2014(2014+1)(2\times 2014+1)}{6}}+\color{#20A900}{2014(2014+1)}\\ \huge&\huge\color{#007fff}{=\boxed{2729146225}} \end{aligned}$