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

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(C) Given the sum of squares: $\sum_{n=1}^{2009} n^2 = (2009)(335)(4019)$ .....$(1)$We need to evaluate the sum $S = \sum_{n=1}^{2009} n(2010 - n)$.Ex

Vedclass · Accepted Answer

$2011$

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(C) Given the sum of squares: $\sum_{n=1}^{2009} n^2 = (2009)(335)(4019)$ .....$(1)$We need to evaluate the sum $S = \sum_{n=1}^{2009} n(2010 - n)$.Expanding the summation: $S = 2010 \sum_{n=1}^{2009} n - \sum_{n=1}^{2009} n^2$.Using the formula for the sum of first $N$ natural numbers,$\sum_{n=1}^{N} n = \frac{N(N+1)}{2}$,where $N = 2009$:$S = 2010 \cdot \frac{2009 \cdot 2010}{2} - (2009)(335)(4019)$.$S = 1005 \cdot 2009 \cdot 2010 - (2009)(335)(4019)$.Since $1005 = 3 \cdot 335$,we have $S = 3 \cdot 335 \cdot 2009 \cdot 2010 - (2009)(335)(4019)$.Factor out $(2009)(335)$: $S = (2009)(335) [3 \cdot 2010 - 4019]$.$S = (2009)(335) [6030 - 4019] = (2009)(335)(2011)$.Comparing this with the given expression $(2009)(335)(x)$,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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