JEE Novice - (12)

It is given that $\space \displaystyle \sum_{i=1}^{2003} {i^2} = (2003)(4007)(334)$ .

$(1)(2003)+(2)(2002)+(3)(2001)+...+(2003)(1) \\ = \displaystyle \sum_{i=1}^{2003} {\left[ i(2004-i) \right]} = 2004 \sum_{i=1}^{2003} {i} - \sum_{i=1}^{2003} {i^2} \\ = 2004\times \dfrac{(2003)(2004)}{2} - (2003)(4007)(334) \\ = (2004)(2003)(1002) - (2003)(4007)(334) \\ = (2004)(2003)(334)(3) - (2003)(4007)(334) \\ = (2003)(334)(6012-4007) \\ = (2003)(334)(2005) \\ = (2003)(334)x$

$\Rightarrow x = \boxed{2005}$

Observe that $\frac{x}{(1-x)^2} = x + 2x + 3x^2 + ...$ .

The expression $1(2003) + 2(2002) + ... + (2003)(1)$ is the coefficient of $x^{2004}$ in the polynomial $\frac{x}{(1-x)^2} \cdot \frac{x}{(1-x)^2} = \frac{x^2}{(1-x)^4}.$

But, $\frac{x^2}{(1-x)^4} = x^2\sum_{n \geq 0} \binom{n+3}{3} x^n$ so the desired coefficient is $\binom{2005}{3}$ . Thus, we have $(2003)(334)x = \binom{2005}{3} = 2005 \cdot 334 \cdot 2003 \implies \boxed{x = 2005}$