The largest perfect square that divides 2014^ | vedclass

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Question

(C) Let $S = 2014^3 - 2013^3 + 2012^3 - 2011^3 + \ldots + 2^3 - 1^3$.We can group the terms as pairs: $(2014^3 - 2013^3) + (2012^3 - 2011^3) + \ldots

Vedclass · Accepted Answer

$1007$

Vedclass · Answer

(C) Let $S = 2014^3 - 2013^3 + 2012^3 - 2011^3 + \ldots + 2^3 - 1^3$.We can group the terms as pairs: $(2014^3 - 2013^3) + (2012^3 - 2011^3) + \ldots + (2^3 - 1^3)$.Using the identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$,and noting that $a - b = 1$ for each pair:$S = (2014^2 + 2014 	imes 2013 + 2013^2) + (2012^2 + 2012 	imes 2011 + 2011^2) + \ldots + (2^2 + 2 	imes 1 + 1^2)$.There are $1007$ such pairs.Alternatively,consider the sum $S = \sum_{k=1}^{1007} ((2k)^3 - (2k-1)^3) = \sum_{k=1}^{1007} (8k^3 - (8k^3 - 12k^2 + 6k - 1)) = \sum_{k=1}^{1007} (12k^2 - 6k + 1)$.$S = 12 \sum_{k=1}^{1007} k^2 - 6 \sum_{k=1}^{1007} k + \sum_{k=1}^{1007} 1$.Using sum formulas $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum k = \frac{n(n+1)}{2}$ with $n = 1007$:$S = 12 	imes \frac{1007(1008)(2015)}{6} - 6 	imes \frac{1007(1008)}{2} + 1007$.$S = 2 	imes 1007 	imes 1008 	imes 2015 - 3 	imes 1007 	imes 1008 + 1007$.$S = 1007 [2 	imes 1008 	imes 2015 - 3 	imes 1008 + 1]$.$S = 1007 [4062360 - 3024 + 1] = 1007 	imes 4059337$.Since $4059337 = 4031 	imes 1007$,we have $S = 1007^2 	imes 4031$.The largest perfect square dividing $S$ is $1007^2$.

The largest perfect square that divides $2014^3 - 2013^3 + 2012^3 - 2011^3 + \ldots + 2^3 - 1^3$ is (in $^2$)

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