waheguru
  • waheguru
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 =)
Discrete Math
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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waheguru
  • waheguru
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KingGeorge
  • KingGeorge
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.
KingGeorge
  • KingGeorge
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.

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