Question

F = 1 + 3¹ + 3² + 3³ + ... + 3¹⁰⁰

^{ai giúp với ạ}

Answer 1

\(F=1+3^1+3^2+...+3^{100}\)

\(\Rightarrow3F=3+3^2+...+3^{101}\)

\(\Rightarrow2F=3F-F=3+3^2+...+3^{101}-1-3^1-...-3^{100}=3^{101}-1\)

\(\Rightarrow F=\dfrac{3^{101}-1}{2}\)

Answer 2

\(3\cdot F=3^1+3^2+...+3^{101}\)

hay \(F=\dfrac{3^{101}-1}{2}\)