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The largest perfect square that divides $2014^{3}-2013^{3}+2012^{3}-2011^{3}+\ldots .+2^{3}-1^{3}$ is
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Verified Answer
The correct answer is:
$1007^{2}$
$S=2014^{3}-2013^{3}+2012^{3}-2011^{3}+\ldots .+2^{3}-1^{3}$
$\sum_{t=1}^{2014} r^{2}+\sum_{t=1}^{107}(2 r-1)(2 r)$
$\mathrm{S}=\sum_{r=1}^{2014} r^{2}+\sum_{r=1}^{1007}\left(4 r^{2}-2 r\right)$
$\mathrm{S}=\frac{(2014)(2015)(4029)}{6}+\frac{4 \times 1007 \times 1008 \times 2015}{6}-\frac{2 \times 1007 \times 1008}{2}=(1007)^{2} \times 4031$
$\sum_{t=1}^{2014} r^{2}+\sum_{t=1}^{107}(2 r-1)(2 r)$
$\mathrm{S}=\sum_{r=1}^{2014} r^{2}+\sum_{r=1}^{1007}\left(4 r^{2}-2 r\right)$
$\mathrm{S}=\frac{(2014)(2015)(4029)}{6}+\frac{4 \times 1007 \times 1008 \times 2015}{6}-\frac{2 \times 1007 \times 1008}{2}=(1007)^{2} \times 4031$
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