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please , give me the reply quickly

$a^{2012} + a^{2013}+.....+a^{3012}=a^{2012}(1+a+a^2+.....+a^{1000})$.
The number in the bracket is odd. So, the highest power of two dividing the whole number is the highest power of two dividing $a^{2012}$. Which is clearly $2012$. So, $n=2012$.

tnx,bro....bt one person said me that the ans will be the sum of powers....my ans is same with u

That person is incorrect.

I would like to add some corrections from "ahmedittihad" 's answer.

According to question, $a = 2^k$ Where, $k$ is any odd natural number

Now, $ a^{2012} + a^{2013} + ...... + a^{3012} $

= $ a^{2012} ( 1 + a + a^{2} +.........+ a^{1000} )$
= $ 2^{2012 k} ( 1 + 2^{k} + 2^{2 k} +.........+ 2^{1000 k} )$

Since, of the two factors only $ 2^{2012 k} $ is even, so maximum power of 2 dividing $a^{2012} = 2^{2012k} $ is the required answer.

Therefore $ n = 2012 k $ Where, $k$ is any odd natural number

Where, $ n = 2012 $ is only one of the possible answers.