$\text{The value of }\left(\begin{array}{c}\text{2008}\\ 3\end{array}\right)-2\left(\begin{array}{c}\text{2008}\\ 4\end{array}\right)+3\left(\begin{array}{c}\text{2008}\\ 5\end{array}\right)-4\left(\begin{array}{c}\text{2008}\\ 6\end{array}\right)+.......2005\left(\begin{array}{c}\text{2008}\\ 2007\end{array}\right)is,where\left(\begin{array}{c}\text{n}\\ r\end{array}\right)=C\left(n,r\right),\text{is}$

$$\begin{gathered} f\left(x\right)=\frac{{\left(1+x\right)}^{2008}}{x^2},f^1\left(-1\right)=0 \\ {f}^1\left(x\right)=\frac{x^2\left(2008\right){\left(1+x\right)}^{2007}-{\left(1+x\right)}^{2008}2x}{x^4} \end{gathered}$$

$f^1\left(-1\right)=0$

$Left hand side of the equation f^1\left(x\right)=\frac{x^2\left(2008\right){\left(1+x\right)}^{2007}-{\left(1+x\right)}^{2008}2x}{x^4}$

$\left(\begin{array}{c}2008\\ 0\end{array}\right)\left(2\right)+\left(\begin{array}{c}2008\\ 1\end{array}\right)\left(-1\right)+\left(\begin{array}{c}2008\\ 2\end{array}\right)\left(0\right)+\left(\begin{array}{c}2008\\ 3\end{array}\right)\left(1\right)+2\left(\begin{array}{c}2008\\ 4\end{array}\right)\left(-1\right)+...-2006\left(\begin{array}{c}2008\\ 2008\end{array}\right)=0$

$The given expression =-2+2008+2006=4012$