a: \(S=3+3^3+3^5+...+3^{2011}+3^{2013}+3^{2015}\)
\(=3+3^2\left(3+3^3+...+3^{2009}+3^{2011}+3^{2013}\right)\)
\(=3+9\left(3+3^3+...+3^{2011}+3^{2013}\right)\)
3 không chia hết cho 9
\(9\cdot\left(3+3^3+...+3^{2011}+3^{2013}\right)⋮9\)
=>\(S=3+9\left(3+3^3+...+3^{2011}+3^{2013}\right)⋮̸9\)
b: \(S=3+3^3+3^5+...+3^{2011}+3^{2013}+3^{2015}\)
\(=\left(3+3^3+3^5+3^7+3^9+3^{11}\right)+\left(3^{13}+3^{15}+3^{17}+3^{19}+3^{21}+3^{23}\right)+...+\left(3^{2005}+3^{2007}+3^{2009}+3^{2011}+3^{2013}+3^{2015}\right)\)
\(=\left(3+3^3+3^5+3^7+3^9+3^{11}\right)+3^{12}\left(3+3^3+...+3^{11}\right)+...+3^{2004}\left(3+3^3+...+3^{11}\right)\)
\(=199290\left(1+3^{12}+...+3^{2004}\right)\)
mà 199290 chia hết cho 70
nên \(S⋮70\)
