HARD
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If (1) (2010) + (2) (2009) + (3) (2008) + ...... + (2010) (1) = (335) (2011) , then the value of must be
(a)2013
(b)2012
(c)2011
(d)2008

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Important Questions on Sequence and Series

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The numbers are distinct ;
are pairwise disjoint.
(Here denotes the number of elements in the set ). Then the maximum possible value of is


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For every natural number . Then is equal to

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has two consecutive integral solutions, then is equal to:


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