NN
17 tháng 3 2020 lúc 20:14
Ta có:
\(A=3^{2000}+...+3^{2012}+3^{2013}⋮3\left(1\right)\)
Lại có:
\(A=3^{2000}+3^{2001}...+3^{2012}+3^{2013}\)
\(\Rightarrow A=\left(3^{2000}+3^{2001}\right)+...+\left(3^{2012}+3^{2013}\right)\)
\(\Rightarrow A=3^{2000}\left(1+3\right)+...+3^{2012}\left(1+3\right)\)
\(\Rightarrow A=3^{2000}.4+...+3^{2012}.4⋮4\left(2\right)\)
Từ (1) và (2)
\(\Rightarrow A=3^{2000}+...+3^{2012}+3^{2013}⋮12\left(đpcm\right)\)