Hard problem I can not prove properly. I managed to concoct a weak proof for the below problem but I would love to see if there are any different ways to solve it.
Find the largest perfect square that divides $2014^3-2013^3+2012^3-2011^3 .... +2^3-1^3$
the options listed are 
a)$1^2$
b)$2^2$
c)$1007^2$
d)$2014^2$
I think the answer is c) but i just got that by calculating the last digit of the sum to check the divisibility. But I doubt that the proof is concrete. Help would be appreciated.
 A: Hint:
$$(2n)^3-(2n-1)^3=3\cdot2^2\cdot n^2-3\cdot2\cdot n+1$$
Put $n=1$ to $2014/2$ and add
A: Hint: Alternatively to lab bhattacharjee's quick solution. 
These are sums that are related to the cubic sum
$$\sum_{k=1}^{n}k^3=\left(\frac{(n+1)n}{2}\right)^2.$$ 
For the even terms you can factor 8 from each term and get a sum from 1^3 to 1007^3. For the odd terms you can add even terms to them to create a complete sum going from 1^3 to 2013^3. 
By using this approach you can get a closed expression for this sum.
A: $$\sum_{k=1}^n((2 k)^3-(2 k-1)^3) = n^2 (4 n+3)$$
in the question the $n\rightarrow \frac{n}{2}$
so the sum will be
$$n^2 (4 n+3)n \rightarrow (\frac{n}{2})^2(2n+3)$$
A: Consider:
\begin{align}
A_{2014} &= \sum_{n=1}^{1007} [(2n)^3 - (2n-1)^3] = \sum_{n=1}^{1007} ((2n) -(2n-1))((2n)^2 + (2n)(2n-1) + (2n-1)^2) \\
&= \sum_{n=1}^{1007} (12n^2 - 6n + 1) \\
&= 4087631519 = 19^{2} \cdot 29^{1} \cdot 53^{2} \cdot 139 = 29 \cdot 139 \cdot (1007)^2.
\end{align}
In comparison: $2014^{2} = 2^2 \cdot 19^2 \cdot 53^2$, $1007^{2} = 19^2 \cdot 53^2$
