What is the largest perfect square that divides $2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3$ I've tried this but didn't get the answer :
Let $S=2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3$
Using $n^3-(n-1)^3 = 3n^2-3n+1$,
\begin{align}
S &= 3(2014^2)-3(2014)+1+3(2012^2)-3(2012)+1+\ldots+3(2^2)-3(2)+1
\\&= 3\left ( 2014(2013)+2012(2011)+2010(2009)+ \ldots+2(1)  \right ) + 1(1007)
\\&= 3\left ( \sum_{n=1}^{1007}2n(2n-1) \right )+1007\\
=& \left ( \sum_{n=1}^{1007}4n^2-\sum_{n=1}^{1007}2n \right )+1007
\\=&\frac{12(1007)(1008)(2015)}{6}-\frac{2(1007)(1008)(3)}{2}+1007
\end{align}
This is divisible by $1007$ but not by $1007^2$ which is the correct answer. Where have I gone wrong ?

\begin{align}&\frac{12(1007)(1008)(2015)}{6}-\frac{2(1007)(1008)(3)}{2}+1007
\\&=1007\left(\frac{12(1008)(2015)}{6}-\frac{2(1008)(3)}{2}+ 1\right)
\\&=1007\left(2(1008)(2015)-(1008)(3)+ 1\right)
\\&=1007\left(2(1007+1)(2015)-(1007+1)(3)+ 1\right)
\\&=1007\left(1007(2(2015)-3)+2(2015)-3+ 1\right)
\\&=1007\left(1007(2(2015)-3)+2(2015)-2\right)
\\&=1007\left(1007(2(2015)-3)+2(2014)\right)
\\&=1007\left(1007(2(2015)-3)+4(1007)\right)
\\&=1007^2(2(2015)-3+4)
\\&=1007^2(4031)\end{align}
