Question 888240
Sum the Series:
1×2+2×3+3×4+4×5+...+2013×2014
<pre>

We will use the formulas: {{{sum((k),k=1,n)=n(n+1)/2}}} and {{{sum((k^2),k=1,n)=n(n+1)(2n+1)/6}}}

1×2+2×3+3×4+4×5+...+2013×2014 = {{{sum((k)(k+1),k=1,2013)}}} = {{{sum((k^2+k),k=1,2013)}}} = {{{sum((k^2),k=1,2013)+sum((k),k=1,2013)}}} =

       {{{2013(2013+1)(2*2013+1)/6+2013(2013+1)/2}}} = 2723058910

Edwin</pre>