\(A=1+2^1+2^2+...+2^{2007}\)
\(\Rightarrow2A=2+2^2+...+2^{2008}\)
\(\Rightarrow2A-A=\left(2+2^2+...+2^{2008}\right)-\left(1+2+...+2^{2007}\right)\)
\(\Rightarrow A=2^{2008}-1\)
\(A=1+3+...+3^7\)
\(\Rightarrow3A=3+3^2+...+3^8\)
\(\Rightarrow3A-A=\left(3+3^2+...+3^8\right)-\left(1+3+...+3^7\right)\)
\(\Rightarrow2A=3^8-1\)
\(\Rightarrow A=\frac{3^8-1}{2}\)
\(A=1+2^1+2^2+2^3+...+2^{2007}\)
\(2A=2+2^2+2^3+...+2^{2008}\)
\(2A-A=\left(2+2^2+2^3+...+2^{2008}\right)-\left(1+2^1+2^2+2^3+...+2^{2007}\right)\)
\(A=2^{2008}-1\)
\(2A=2^{2009}-2\)
b) \(A=1+3^1+3^2+3^3+...+3^7\)
\(3A=3+3^2+3^3+3^4+...+3^8\)
\(3A-A=\left(3+3^2+3^3+...+3^8\right)-\left(1+3^1+3^2+3^3+...+3^7\right)\)
\(2A=3^8-1\)
\(\Rightarrow A=\frac{3^8-1}{2}\left(\text{đpcm}\right)\)
Chúc bạn học tốt !!!
vãi đái, khó thế.
