## waheguru Group Title this is a question from a past math contest but I do not understand how we even get near to finding the ansnwe. Can some one please answer this step by step. Thanks =) one year ago one year ago

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1. waheguru Group Title

2. KingGeorge Group Title

So here's the way I looked at it. The pattern is simply $n^2-(n+1)^2-(n+2)^2+(n+3)^2+...$Starting at $$n=1$$, and stopping as soon as we hit 2011. That expression simplifies as $n^2-(n+1)^2-(n+2)^2+(n+3)^2=4.$That's right. Just 4.

3. KingGeorge Group Title

Then, $$2011=502\cdot 4+3$$. So we have 502 subsequences that look like $n^2-(n+1)^2-(n+2)^2+(n+3)^2,$and then we finish with $2009^2-2010^2-2011^2.$So the total sum would be$502*4+2009^2-2010^2-2011^2=-4046132$which is choice E.