\(S=1+3^2+3^3+...+3^{2020}\)
\(3S=3\left(1+3^2+3^3+...+3^{2020}\right)\)
\(3S=3+3^3+3^4+...+3^{2021}\)
\(2S=\left(3^{2021}+3\right)-\left(1+3^2\right)=3^{2021}-11\)
\(S=\frac{3^{2021}-11}{2}\)
\(S=\left(3^{2020}.3-11\right):2\)
\(S=\left[\left(...3\right)-11\right]:2\)
\(S=\left(...2\right):2=\left(...1\right)VS\left(...6\right)\)
\(1+3^2+3^4+3^6+...+3^{2020}\)
\(\Rightarrow\)\(9S=3+3^2+3^4+3^6+...+3^{2020}\)
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\(S=1+3^2+3^4+3^6+...+3^{2020}\)
Bằng\(8S=3^{2020}-1\)
\(\Rightarrow\)\(S=\left(3^{2020}-1\right):8\)
Ta có:\(S=\left(3^{2020}-1\right):3\)
\(=\)[ (...1)-1]:3
\(=\)...0:3
\(=\)...0
Vậy chữ số tận cùng của S là 0
Cho mik mấy k nha
