e: \(E=1+3+3^2+3^3+...+3^{119}\)
\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{117}+3^{118}+3^{119}\right)\)
\(=13+3^3\cdot\left(1+3+3^2\right)+...+3^{117}\left(1+3+3^2\right)\)
\(=13\left(1+3^3+...+3^{117}\right)⋮13\)
f: \(F=10^{28}+8\)
\(\left\{{}\begin{matrix}10^{28}=10^3\cdot10^{25}=8\cdot125\cdot10^{25}⋮8\\8⋮8\end{matrix}\right.\)
=>\(F=10^{28}+8⋮8\)
\(10^{28}+8=10...08\)
Tổng các chữ số của F là 1+0+...+0+8=9
=>F chia hết cho 9
mà F chia hết cho 8
nên F chia hết cho BCNN(9;8)=72
g: \(G=8^8+2^{20}\)
\(=\left(2^3\right)^8+2^{20}\)
\(=2^{24}+2^{20}\)
\(=2^{20}\left(2^4+1\right)\)
\(=2^{10}\cdot17⋮17\)
h: \(H=2+2^2+2^3+...+2^{60}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)
\(=15\left(2+2^5+...+2^{57}\right)⋮15\)
=>\(H⋮3\)
\(H=2+2^2+2^3+...+2^{60}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=7\left(2+2^4+...+2^{58}\right)⋮7\)
