Đặt \(A=1+3+3^2+3^3+...+3^{2017}\)
\(3A=3\left(1+3+3^2+3^3+...+3^{2017}\right)\)
\(=3+3^2+3^3+3^4+...+3^{2018}\)\(3A-A=\left(3+3^2+3^3+3^4+...+3^{2018}\right)-\left(1+3+3^2+3^3+...+3^{2017}\right)\)\(2A=3^{2018}-1\Rightarrow A=\frac{3^{2018}-1}{2}\)
Vậy \(A=\frac{3^{2018}-1}{2}\)
gọi bieu thuc tren la A
A= 1+3+3^2+..+3^2017
3A= 3.(1+3+362+..+3^2017)
3A=3+3^2+3^3+...+3^2018
3A - A= (3+ 3^2+3^3+...+3^2018) - (1+3+3^2+...+3^2017)
2A= 3^2018 - 1
=> A= \(\frac{3^{2018}-1}{2}\)
mk k biết có đúng k nữa, bạn thông cảm cho mk nhé
