Cho $f\left(x\right)=1+x^2+x^4+x^6+...+x^{2020}$. Tính f(1); f(-1).

Đáp án cần chọn là đáp án C.

Thay x = -1 vào $P\left(x\right)=100x^{100}+99x^{99}+98x^{98}+...+2x^2+x$ ta được:

$$\begin{gathered} P\left(x\right)=100{\left(-1\right)}^{100}+99{\left(-1\right)}^{99}+98{\left(-1\right)}^{98}+...+2{\left(-1\right)}^2+\left(-1\right) \\ =100-99+98-97+...+2-1 \\ =1+1+...+1 \\ =50.1=50 \end{gathered}$$

Vậy $P(-1)=50$.

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.