Bài1:
\(A=3+3^3+3^5+...+3^{1991}\)
=\(\left(3+3^3+3^5\right)+...+\left(3^{1989}+3^{1990}+3^{1991}\right)\)
=\(3\left(1+3^2+3^4\right)+...+3^{1989}\left(1+3^2+3^4\right)\)
=\(91\left(3+...+3^{1989}\right)\)
Vì \(91\left(3+...+3^{1989}\right)\) chia hết cho 13 nên Achia hết cho 13 (đpcm)
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