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

Đáp án:

$\dfrac{4^{2019}+3^{2019}}{7.4^{2018}}.$

Giải thích các bước giải:

$A=1-\dfrac{3}{4}+\left(\dfrac{3}{4}\right)^2-\left(\dfrac{3}{4}\right)^3+\left(\dfrac{3}{4}\right)^4-\dots-\left(\dfrac{3}{4}\right)^{2017}+\left(\dfrac{3}{4}\right)^{2018}\\ \dfrac{3}{4}A=\dfrac{3}{4}-\left(\dfrac{3}{4}\right)^2+\left(\dfrac{3}{4}\right)^3-\left(\dfrac{3}{4}\right)^4+\left(\dfrac{3}{4}\right)^5-\dots-\left(\dfrac{3}{4}\right)^{2018}+\left(\dfrac{3}{4}\right)^{2019}\\ A+\dfrac{3}{4}A=1+\left(\dfrac{3}{4}\right)^{2019}\\ \Leftrightarrow \dfrac{7}{4}A=1+\left(\dfrac{3}{4}\right)^{2019}\\ \Leftrightarrow A=\dfrac{4}{7}+\dfrac{4}{7}.\left(\dfrac{3}{4}\right)^{2019}\\ \Leftrightarrow A=\dfrac{4}{7}+\dfrac{4}{7}.\dfrac{3^{2019}}{4^{2019}}\\ \Leftrightarrow A=\dfrac{4}{7}+\dfrac{3^{2019}}{7.4^{2018}}\\ \Leftrightarrow A=\dfrac{4^{2019}+3^{2019}}{7.4^{2018}}.$