The largest perfect square that divides $2014^{3} - 2013^{3} + 2012^{3} - 2011^{3} + . . . . + 2^{3} - 1^{3}$ is.

1^{2}

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2^{2}

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1007^{2}

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2014^{2}

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Solution

The correct option is D $1007^{2}$
$S = 2014^{3} - 2013^{3} + 2012^{3} - 2011^{3} + . . . . + 2^{3} - 1^{3}$
$2014 \sum r = 1 r^{2} = 1007 \sum r = 1 (2 r - 1) (2 r)$
$S = 2014 \sum r = 1 r^{2} + 1007 \sum r = 1 (4 r^{2} - 2 r)$
$S = \frac{(2014) (2015) (4029)}{6} + \frac{4 \times 1007 \times 1008 \times 2015}{6} - \frac{2 \times 1007 \times 1008}{2} = (1007)^{2} \times 4031$

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Similar questions

(2012)^{3} + (2013)^{3} + (2014)^{3} - 3 \times 2012 \times 2013 \times 2014

= (. . . . . . . . .) \times {\begin{matrix} (2012)^{2} + (2013)^{2} + (2014)^{2} - 2012 \times 2013 - 2013 \times 2014 - 2014 \times 2012 \end{matrix}}

\frac{3 \times 1^{3}}{1^{2}} + \frac{5 \times (1^{3} + 2^{3})}{1^{2} + 2^{2}} + \frac{7 \times (1^{3} + 2^{3} + 3^{3})}{1^{2} + 2^{2} + 3^{2}} + \dots

10^{th}

\frac{3 \times 1^{3}}{1^{2}} + \frac{5 \times (1^{3} + 2^{3})}{1^{2} + 2^{2}} + \frac{7 \times (1^{3} + 2^{3} + 3^{3})}{1^{2} + 2^{2} + 3^{2}} + \dots upto 10 terms

The ratio obtained when the largest 4-digit perfect square number divided by the smallest 4-digit perfect square number will be

S = \sqrt{1 + \frac{1}{1^{2}} + \frac{1}{2^{2}}} + \sqrt{1 + + \frac{1}{2^{2}} + \frac{1}{3^{2}}} + . . . + \sqrt{1 + + \frac{1}{{(2014)}^{2}} + \frac{1}{{(2015)}^{2}}}

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