Có: \(B=1+4+4^2+...+4^{2009}\)
=> \(4.B=4.\left(1+4+4^2+...+4^{2019}\right)\)
\(4B=4+4^2+4^3+...+4^{2020}\)
=> \(4B-B=\left(4+4^2+4^3+...+4^{2020}\right)-\left(1+4+4^2+...+4^{2019}\right)\)
\(3B=\left(4-4\right)+\left(4^2-4^2\right)+...+\left(4^{2019}-4^{2019}\right)+\left(4^{2020}-1\right)\)
\(3B=4^{2020}-1\)
=> \(3B+1=4^{2020}-1+1\)
\(3B+1=4^{2020}\)
Vậy 3B + 1 là lũy thừa của 4.
\(B=1+4+4^2+......+4^{2019}\)
\(\Rightarrow4B=4+4^2+4^3+.......+4^{2020}\)
\(\Rightarrow4B-B=3B=4^{2020}-1\)
Ta có: \(3B+1=4^{2020}-1+1=4^{2020}\)là lũy thừa của 4 ( đpcm )