2 divides a, but 4 doesn't .
so a is form as $a=2(2k+1)$
so,
$a^{2012}+a^{2013}+a^{2014}+.........+a^{3012}$
$=\left \{ 2(2k+1) \right \}^{2012}+\left \{ 2(2k+1) \right \}^{2013}+\left \{ 2(2k+1) \right \}^{2014}+.........+\left \{ 2(2k+1) \right \}^{3012} $
$=2^{2012}(2k+1)^{2012}+2^{2013}(2k+a)^{2013}+2^{2014}(2k+1)^{2014}+.........+2^{3012}(2k+1)^{3012}$
$=2^{2012}\left \{ (2k+1)^{2012}+2(2k+1)^{2013}+2^{2}(2k+1)^{2014}+.........+2^{1000}(2k+1)^{3012} \right \}$
now let $\left \{ (2k+1)^{2012}+2(2k+1)^{2013}+2^{2}(2k+1)^{2014}+.........+2^{1000}(2k+1)^{3012} \right \}=x$
and here x is a odd number.
so,$2^{n}$ does not divide x.
that imply $2^{n}$ must divide $2^{2012}$.
so,the maxium value of n is 2012 for which $2^{n}$ divides $a^{2012}+a^{2013}+a^{2014}+.........+a^{3012}$
n=2012 ...(ans:)...