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 ?

In general you obtain, for $n^3-(n-1)^3+\cdots + 2^3-1^3$ and $n$ even the formula
$$
3\left(\sum_{k=1}^{n/2} 2k(2k-1)\right)+\frac{n}{2}.
$$
Now it is easier to see that this is divisible by $(\frac{n}{2})^2$; and you can test this first for $n=2,4,6,\ldots$ before dealing with $n=2014$. In fact, your computation is correct, except for the last step, where you did not realize how to split of the factor $(\frac{n}{2})^2$. Working with general $n$, you do not need a calculator.
