Question

rút gọn:A=3+3²+3³+...+3²⁰⁰⁴

giúp mình câu này 

Answer 1

\(A=3+3^2+3^3+...+3^{2004}\)

\(\Rightarrow3A=3\left(3+3^2+3^3+...+3^{2004}\right)\)

\(\Rightarrow3A=3^2+3^3+3^4+...+3^{2005}\)

\(\Rightarrow3A-A=\left(3^2+3^3+3^4+...+3^{2005}\right)-\left(3+3^2+3^3+3^4+...+3^{2004}\right)\)

\(\Rightarrow2A=\left(3^2-3^2\right)+\left(3^3-3^3\right)+\left(3^4-3^4\right)+...+\left(3^{2004}-3^{2004}\right)+\left(3^{2005}-3\right)\)

\(\Rightarrow2A=3^{2005}-3\)

\(\Rightarrow A=\dfrac{3^{2005}+3}{2}\)