Giải thích các bước giải:
Ta có:
$f(x)=x^{99}-3000x^{98}+3000x^{97}-3000x^{96}+...-3000x^2+3000x-1$
$\to xf(x)=x^{100}-3000x^{99}+3000x^{98}-3000x^{97}+...-3000x^3+3000x^2-x$
$\to f(x)+xf(x)=x^{100}-2999x^{99}+2999x-1$
$\to (1+x)f(x)=x^{100}-2999x^{99}+2999x-1$
$\to f(x)=\dfrac{x^{100}-2999x^{99}+2999x-1}{x+1}$
$\to f(2009)=\dfrac{2009^{100}-2999\cdot 2009^{99}+2999\cdot 2009-1}{2009+1}$
$\to f(2009)=\dfrac{2009^{100}-2999\cdot 2009^{99}+2999\cdot 2009-1}{2010}$
