Hard problem I can not prove properly.

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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: $2014^{2} = 2^2 \cdot 19^2 \cdot 53^2$, $1007^{2} = 19^2 \cdot 53^2$

Answer 4

$$\sum_{k=1}^n((2 k)^3-(2 k-1)^3) = n^2 (4 n+3)$$

in the question the $n\rightarrow \frac{n}{2}$ so the sum will be $$n^2 (4 n+3)n \rightarrow (\frac{n}{2})^2(2n+3)$$