If 1^2 + 2^2 + 3^2 + dots + 2009^2 = (2009)(3 | vedclass

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(C) The given sum is $S = \sum_{n=1}^{2009} n(2010 - n)$.This can be written as $S = 2010 \sum_{n=1}^{2009} n - \sum_{n=1}^{2009} n^2$.We know $\sum_{

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$2011$

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(C) The given sum is $S = \sum_{n=1}^{2009} n(2010 - n)$.This can be written as $S = 2010 \sum_{n=1}^{2009} n - \sum_{n=1}^{2009} n^2$.We know $\sum_{n=1}^{N} n = \frac{N(N+1)}{2}$,so $\sum_{n=1}^{2009} n = \frac{2009 	imes 2010}{2} = 2009 	imes 1005$.Thus,$S = 2010(2009 	imes 1005) - (2009)(335)(4019)$.$S = (2009)(2010 	imes 1005) - (2009)(335)(4019)$.$S = (2009)(335) 	imes [3 	imes 2010 - 4019]$.$S = (2009)(335) 	imes [6030 - 4019]$.$S = (2009)(335) 	imes 2011$.Comparing this with $(2009)(335)(x)$,we get $x = 2011$.

If $1^2 + 2^2 + 3^2 + \dots + 2009^2 = (2009)(335)(4019)$ and $(1)(2009) + 2(2008) + 3(2007) + \dots + 2009(1) = (2009)(335)(x)$,then $x$ is equal to:

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