# Question 60895

Oct 31, 2016

${1007}^{2}$

#### Explanation:

Calling $n = 2014$ we have

2014³ - 2013³ + 2012³ - 2011³ + ... + 2³- 1³= sum_(k=1)^(n/2)(2k)^3-sum_(k=1)^(n/2)(2k-1)^3 #

but

${\sum}_{k = 1}^{n} {k}^{3} = {\left(\frac{n \left(n + 1\right)}{2}\right)}^{2}$

Now

${\sum}_{k = 1}^{\frac{n}{2}} {\left(2 k\right)}^{3} = {2}^{3} {\sum}_{k = 1}^{\frac{n}{2}} {k}^{3} = {2}^{3} {\left(\frac{\frac{n}{2} \left(\frac{n}{2} + 1\right)}{2}\right)}^{2}$

and

${\sum}_{k = 1}^{\frac{n}{2}} {\left(2 k - 1\right)}^{3} = {\left(\frac{n \left(n + 1\right)}{2}\right)}^{2} - {2}^{3} {\left(\frac{\frac{n}{2} \left(\frac{n}{2} + 1\right)}{2}\right)}^{2}$

and

${\sum}_{k = 1}^{\frac{n}{2}} {\left(2 k\right)}^{3} - {\sum}_{k = 1}^{\frac{n}{2}} {\left(2 k - 1\right)}^{3} = 2 \times {2}^{3} {\left(\frac{\frac{n}{2} \left(\frac{n}{2} + 1\right)}{2}\right)}^{2} - {\left(\frac{n \left(n + 1\right)}{2}\right)}^{2}$

$= {\left(\frac{n}{2} \left(n + 2\right)\right)}^{2} - {\left(\frac{n}{2} \left(n + 1\right)\right)}^{2} = {\left(\frac{n}{2}\right)}^{2} \left({\left(n + 2\right)}^{2} - {\left(n + 1\right)}^{2}\right) =$
$= {\left(\frac{n}{2}\right)}^{2} \left(2 n + 3\right)$

in this case

$2 n + 3 = 4031 = 29 \times 139$

so the largest perfect square is ${\left(\frac{2014}{2}\right)}^{2} = {1007}^{2}$