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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