Question

1+4^1 +4^2+.....+4^2022

ai giúp mình với

 

Answer 1

\(4\cdot A=4+4^2+...+4^{2023}\)

\(\Leftrightarrow3\cdot A=4^{2022}-1\)

hay \(A=\dfrac{4^{2022}-1}{3}\)

Answer 2

1 + 4¹ + 4² + .... + 4²⁰²²

Đặt : 

A = 1 + 4¹ + 4² + .... + 4²⁰²²

4A = 4 + 4² + 4³ + ..... + 4²⁰²³

4A - A = ( 4 + 4² + 4³ + ..... + 4²⁰²³ ) - ( 1 + 4¹ + 4² + .... + 4²⁰²² )

3A = 4²⁰²³ - 1

A = \(\dfrac{4^{2023}-1}{3}\)