If $1^{2} + 2^{2} + 3^{2} + . . . + 2009^{2} = (2009) (4019) (335) a n d (1) (2009) + (2) (2008) + (3) (2007) + . . . + (2009) (1) = (2009) (335) (x),$ , then x equals

2011

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2009

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2008

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2007

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Solution

The correct option is A 2011
Solution: $(1) (2009) + (2) (2008) + (3) (2007) + \dots . . + (2009) (1)$
$T_{n} = n {2009 - (n - 1)} = n (2009 - n + 1)$
$= n (2010 - n)$
$= 2010 n - n^{2}$
$\sum_{n = 1}^{2009} T_{n} = \sum_{n = 1}^{2009} (2010) n - \sum_{n = 1}^{2009} n^{2}$
$= 2010 \sum_{n = 1}^{2009} n - (2009) (4019) (335)$
$= (2009) {(1005) (2010) - (4019) (335)}$
$= (2009) (335) {(3) (2010) - 4019}$
$= (2009) (335) {6030 - 4019}$
$= (2009) (335) (2011)$
therefore $x = 2011$
Answer: option (A).

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