Ta có:
\(S = 1 + 3 + 3^2 + ... + 3^2016\)
\(3S = 3 . ( 1 + 3 + 3^2 + ... + 3^{2017})\)\(\)
\(3S = 3 + 3^2 +3^3 +... + 3^{2018}\)
\(3S-S=(3+3^2+3^3+...+3^{2018})-(1+3+3^2+...+3^{2017})\)
\(2S = 3^{2018} - 1\)
\(S=\dfrac{3^{2018}-1}{2}\)
Vậy \(S=\dfrac{3^{2018}-1}{2}\)
S = 1 + 3 + 3^2 + 3^3 +...+ 3^2017
S . (3 - 1) = 3 - 1 + 3^2 - 3 + 3^3 - 3^2 + ... + 3^2018 - 3^2017
S = (3^101 - 1) / 2
\(S=1+3+3^2+3^3+...+3^{2017}\)
\(3S=3\left(1+3+3^2+3^3+...+3^{2017}\right)\)
\(3S=3+3^2+3^3+3^4+...+3^{2018}\)
\(3S-S=\left(3+3^2+3^3+3^4+...+3^{2018}\right)-\left(1+3+3^2+3^3+...+3^{2017}\right)\)
\(2S=3^{2018}-1\)
\(S=\dfrac{3^{2018}-1}{2}\)
