If 1^2+2^2+3^2++2003^2=(2003)(4007)(334)...

If `1^2+2^2+3^2++2003^2=(2003)(4007)(334)` and `(1)(2003)+(2)(2002)+(3)(2001)+...+(2003)(1)=(2003)(334)(x),` the ` nx` equals

2005

2004

2003

2001

Text Solution

Verified by Experts

The correct Answer is:

`S=(1)(2003)+(2)(2002)+(3)(2001)+..+(2003)(1)`
`=sum_(r=1)^(2003)r(2003-(r-1))`
`=sum_(r=1)^(2003)r(2004-r)`
`=sum_(r=1)^(2003)r(2004-r)`
`sum_(r=1)^(2003)2004r-sum_(r=1)^(2003)r^(2)`
`=(2004xx2003xx2004)/2-2003xx4007xx334`
`=2003xx334xx(6012-4007)`
`=2003xx334xx2005`
Hence,x=2005.

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