Tính giá trị biểu thức A = 1- 3/4+ (3/4)^2- (3/4)^3+ (3/4)^4-....-(3/4)^2001+ (3/4)^2022

Hướng dẫn giải

Đáp án đúng là: A

A = $1-\frac{3}{4}+{\left(\frac{3}{4}\right)}^2-{\left(\frac{3}{4}\right)}^3+{\left(\frac{3}{4}\right)}^4-\mathrm{...}-{\left(\frac{3}{4}\right)}^{2021}+{\left(\frac{3}{4}\right)}^{2022}$

 $\frac{3}{4}\text{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)}^5-\mathrm{...}-{\left(\frac{3}{4}\right)}^{2022}+{\left(\frac{3}{4}\right)}^{2023}$

$\frac{3}{4}\text{A + A =}\left[\frac{3}{4}-{\left(\frac{3}{4}\right)}^2+{\left(\frac{3}{4}\right)}^3-{\left(\frac{3}{4}\right)}^4+{\left(\frac{3}{4}\right)}^5-\mathrm{...}-{\left(\frac{3}{4}\right)}^{2022}+{\left(\frac{3}{4}\right)}^{2023}\right]$+$\left[1-\frac{3}{4}+{\left(\frac{3}{4}\right)}^2-{\left(\frac{3}{4}\right)}^3+{\left(\frac{3}{4}\right)}^4-\mathrm{...}-{\left(\frac{3}{4}\right)}^{2021}+{\left(\frac{3}{4}\right)}^{2022}\right]$

$\frac{7}{4}\text{A = }$$1+{\left(\frac{3}{4}\right)}^{2023}$

A = $\left[1+{\left(\frac{3}{4}\right)}^{2023}\right]:\frac{7}{4}=\left[1+{\left(\frac{3}{4}\right)}^{2023}\right].\frac{4}{7}=\frac{4}{7}+\frac{4}{7}.\frac{3}{4}.{\left(\frac{3}{4}\right)}^{2022}=\frac{4}{7}+\frac{3}{7}.{\left(\frac{3}{4}\right)}^{2022}$

Vậy đáp án đúng là A.