a) \(B=3+3^2+3^3+...+3^{30}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{28}+3^{29}+3^{30}\right)\)
\(=39+3^3\cdot\left(3+3^2+3^3\right)+...+3^{27}\cdot\left(3+3^2+3^3\right)\)
\(=39+3^3+39+...+3^{27}\cdot39\)
\(=39\cdot\left(1+3^3+...+3^{27}\right)\)
Vậy B chia hết cho 39
b) Ta có:
\(3^{n+2}+3^n\)
\(=3^n\cdot\left(3^2+1\right)\)
\(=3^n\cdot\left(9+1\right)\)
\(=10\cdot3^n\)
Vậy \(3^{n+2}+3^n\) chia hết cho 10