Ta có: \(A=1+4+....+4^{50}\)
\(\Rightarrow4A=4+4^2+...+4^{51}\)
\(\Rightarrow4A-A=\left(4+4^2+...+4^{51}\right)-\left(1+4+..+4^{50}\right)\)
\(\Rightarrow3A=4^{51}-1\)
\(\Rightarrow A=\frac{4^{51}-1}{3}\)
\(\Rightarrow A=\frac{...4-1}{3}\)
\(\Rightarrow A=...1\)
Vậy A tận cùng là 1