Ta có :
M = \(2+2^2+2^3+2^4+............+2^{60}\) ( 60 số hạng)
M = (\(2^{60}+2^{59}+2^{58}+2^{57}\))+............+(\(2^4+2^3+2^2+2\)) ( 15 nhóm)
M = \(2^{58}\)(\(2^3+2^2+2+1\))+..............+\(2^2\left(2^3+2^2+2+1\right)\)
M = \(2^{58}.15\) +.............\(2^2.15\)
=>M chia hết cho 15
=> đpcm
Ta có
A=2(1+2)+2^3(1+2)+..............+2^59(1+2)
A=3(2+2^3+2^5+........+2^59)\(\Rightarrow\)\(A⋮3\left(1\right)\)
Ta có :
A=2(1+2+2^2)+2^4(1+2+2^2)+...........+2^58((1+2+2^2)
A=7(2+2^4+2^7+..........+2^58) \(\Rightarrow A⋮7\left(2\right)\)
Từ (1) và (2)\(\Rightarrow\)\(A⋮21\)
M=(2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8)+....+(2^57+2^58+2^59+2^60) chia hết cho15
30+ 2^4.(2+2^2+2^3+2^4).....
30.2^4.30....
Vi 30 chia hết cho 15 suy ra cả biêu thức sẽ chia hêt cho 5
\(Q=3+3^3+3^5+3^7+...+3^{1991}\)
\(Q=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+...+\left(3^{1987}+3^{1989}+3^{1991}\right)\)
Q=3(1+9+81)+3^2(1+9+81)+...+3^1987(1+9+81)
\(Q=3.91+3^7.91+...+3^{1987}.91\)
\(Q=3.13.7+3^7.13.7+...+3^{1987}.13.7\)
\(Q=13.\left(3.7+3^7.7+...+3^{1987}.7\right)\)
\(\Rightarrow Q⋮13\)
\(Q=3+3^3+3^5+3^7+...+3^{1991}\)
\(Q=\left(3+3^3+3^5+3^7\right)+...+\left(3^{1985}+3^{1987}+3^{1989}+3^{1991}\right)\)
\(Q=3\left(1+3^3+3^4+3^6\right)+...+3^{1985}\left(1+3+3^2+3^4+3^6\right)\)
\(Q=3.820+...+3^{1985}.820\)
\(Q=3.20.41+...+3^{1985}.20.41\)
\(Q=41.\left(3.20+...+3^{1985}.20\right)\)
\(\Rightarrow Q⋮41\)