Question
Thu June 07, 2012 By:
 

if n+1= (2010) whole square + (2011) whole square the find root under 2n+1

Expert Reply
Fri June 08, 2012
n + 1 = (2010)2 + (2011)2
n = (2010)2 + (2011)2 - 1
n = (2010)2 + (2011 - 1)(2011 + 1)
n = (2010)2 + (2010)(2012)
n = 2010[2010 + 2012] = (2010) (4022) = 2(2010)(2011)
 
Thus, we have:
 
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