Cho f(x)=(x^99)-101(x^98)+101(x^97)-101(x^96)+...+100x-1. Tính f(100).

* Hướng dẫn giải

Đáp án cần chọn là B.

Ta có:

$$\begin{gathered} f\left(x\right)=x^{99}-101x^{98}+101x^{97}-101x^{96}+...+101x-1 \\ =x^{99}-\left(100+1\right)x^{98}+\left(100+1\right)x^{97}-\left(100+1\right)x^{96}+...-\left(100+1\right)x^2+\left(100+1\right)x-1 \\ =x^{99}-100x^{98}-x^{98}+100x^{97}+...-100x^2-x^2+100x+x-1 \\ =\left(x^{99}-100x^{98}\right)-\left(x^{98}-100x^{97}\right)+...-\left(x^2-100x\right)+x-1 \end{gathered}$$

Thay x = 100 vào f(x) ta được:

$$\begin{gathered} f\left(100\right)=\left({100}^{99}-100.{100}^{98}\right)-\left({100}^{98}-100.{100}^{97}\right)+...-\left({100}^2-100.100\right)+100-1 \\ =\left({100}^{99}-{100}^{99}\right)-\left({100}^{98}-{100}^{98}\right)+...-\left({100}^2-{100}^2\right)+99 \\ =99 \end{gathered}$$

Vậy f(100) = 99.