Question

Cho S = \(3^0\)\(+3^2\)\(+3^4\)\(+....+3^{2016}\). Tính S

Answer 1

\(9S=3^2+3^4+...+3^{2018}\)

=>\(8S=3^{2018}-1\)

hay \(S=\dfrac{3^{2018}-1}{8}\)

Answer 2

9S=3²+3⁴+...+3²⁰¹⁸

9S-S=(3²+3⁴+...+3²⁰¹⁸)-(3⁰+3²+...+3²⁰¹⁶)

8S=3²⁰¹⁸-1

S=\(\dfrac{3^{2018}-1}{8}\)