Cho A = 1 - 3/4 + (3/4)^2 - (3/4)^3 + (3/4)^4 +...- (3/4)^2017 + (3/4)^2018

* Hướng dẫn giải

Đáp án A

$$\begin{gathered} A=1-\frac{3}{4}+{\left(\frac{3}{4}\right)}^2-{\left(\frac{3}{4}\right)}^3+{\left(\frac{3}{4}\right)}^4-...-{\left(\frac{3}{4}\right)}^{2017}+{\left(\frac{3}{4}\right)}^{2018} \\ \Rightarrow \frac{3}{4}A=\frac{3}{4}-{\left(\frac{3}{4}\right)}^2+{\left(\frac{3}{4}\right)}^3-{\left(\frac{3}{4}\right)}^4+...+{\left(\frac{3}{4}\right)}^{2017}-{\left(\frac{3}{4}\right)}^{2018}+{\left(\frac{3}{4}\right)}^{2019} \\ \Rightarrow A+\frac{3}{4}A=1+{\left(\frac{3}{4}\right)}^{2019} \\ \Rightarrow \left(1+\frac{3}{4}\right)A=1+{\left(\frac{3}{4}\right)}^{2019} \\ \Rightarrow \frac{7}{4}A=1+{\left(\frac{3}{4}\right)}^{2019} \\ \Rightarrow A=\left[1+{\left(\frac{3}{4}\right)}^{2019}\right]:\frac{7}{4}=\left[1+{\left(\frac{3}{4}\right)}^{2019}\right].\frac{}{7} \end{gathered}$$

Suy ra A > 1 (1)

Vì ${\left(\frac{3}{4}\right)}^{2019}<\frac{3}{4}\Rightarrow A<\left(1+\frac{3}{4}\right).\frac{4}{7}=1 \left(2\right)$

Vậy A không phải là số nguyên