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Tuesday, September 16, 2008

Q) If 12 + 22 + 32 + ..... + 20032 = (2003)(4007)(334) and (1)(2003)+(2)(2002)+(3)(2001)+....(2003)(1) = (2003)(334)(x), then find x .


A)


So for 2003 terms the summation is = (2003)(2004)(4007) / 6 = (2003)(4007)(334)



Now,


(1)(2003)+(2)(2002)+(3)(2001)+....(2003)(1) =


= [ 2004(2003)(2003 + 1) / 2 ] - [ (2003)(4007)(334) ]


= [ (1002)(2003)(2004) ] - [ (2003)(4007)(334) ]


= 3 [(334)(2003)(2004) ] - [ (2003)(4007)(334)]


= (2003)(334)( 3*2004 - 4007)


= (2003)(334)(2005)



So the answer is 2005.


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