Question

Cho \(A=1-\dfrac{3}{4}+\left(\dfrac{3}{4}\right)^2-\left(\dfrac{3}{4}\right)^3+...-\left(\dfrac{3}{4}\right)^{2019}+\left(\dfrac{3}{4}\right)^{2020}\)

 CM: A không là số nguyên .

Answer 1

Lời giải:

$A=1-\frac{3}{4}+(\frac{3}{4})^2-(\frac{3}{4})^3+....-(\frac{3}{4})^{2019}+(\frac{3}{4})^{2020}$
$\frac{3}{4}A=\frac{3}{4}-(\frac{3}{4})^2+(\frac{3}{4})^3-(\frac{3}{4})^4+...-(\frac{3}{4})^{2020}+(\frac{3}{4})^{2021}$

$\Rightarrow A+\frac{3}{4}A=1+(\frac{3}{4})^{2021}$

$\Rightarrow \frac{7}{4}A=1+(\frac{3}{4})^{2021}$

$\Rightarrow A=\frac{4}{7}+\frac{4}{7}.(\frac{3}{4})^{2021}$

$=\frac{4^{2021}+3^{2021}}{4^{2020}.7}$

Hiển nhiên $4^{2021}+3^{2021}\not\vdots 4^{2020}$ nên $A$ không là số nguyên.