Question

If 12+22+32+.......+20032 = (2003)(4007)(334) and (1)(2003) + (2)(2002) + (3)(2001) + .... + (2003)(1) = (2003)(334)(x), then the value of x is

Answer 1

 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.