\(S=3+3^2+3^3+...+3^{100}\)
\(\Leftrightarrow3S=3^2+3^3+3^4+...+3^{101}\)
\(\Leftrightarrow3S-S=\left(3^2+3^3+3^4+...+3^{101}\right)-\left(3+3^2+3^3+...+3^{100}\right)\)
\(\Leftrightarrow2S=3^{101}-3\)
\(\Leftrightarrow S=\frac{3^{101}-3}{2}\)
Ta thấy : \(S=\frac{3^{101}-3}{2}=\frac{\left(3^4\right)^{25}.3-3}{2}=\frac{\overline{...1}.3-3}{2}=\frac{\overline{...3}-3}{2}=\frac{\overline{...0}}{2}=\overline{...0}\)
Vậy chữ số cuối cùng của S là 0