Question

M=4+4²+4³+ ........ + 4²⁰²⁰+ 4²⁰²¹

Tính M

Answer 1

Ta có: \(4M=4^2+4^3+4^4+...+4^{2021}+4^{2022}\)

  \(\Rightarrow4M-M=4^2+4^3+4^4+...+4^{2021}+4^{2022}-\left(4+4^2+4^3+...+4^{2020}+4^{2021}\right)\)

  \(\Leftrightarrow3M=4^{2022}-4\)

   \(\Leftrightarrow M=\dfrac{4^{2022}-4}{3}\)

Answer 2

\(4M=4^2+4^3+...+4^{2022}\)

\(\Leftrightarrow M=\dfrac{4^{2022}-4}{3}\)