a) \(A=3+3^2+3^3+...+3^{60}\)
Vì \(3⋮3;3^2⋮3;3^3⋮3;...;3^{60}⋮3\)
\(\Rightarrow3+3^2+3^3+...+3^{60}⋮3\\ \Rightarrow A⋮3\)
b) \(A=3+3^2+3^3+...+3^{60}\\ =\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{59}+3^{60}\right)\\ =3\left(1+3\right)+3^3\left(1+3\right)+...+3^{59}\left(1+3\right)\\ =\left(1+3\right)\left(3+3^3+...+5^{59}\right)\\ =4\left(3+3^3+...+5^{59}\right)⋮4\)
a. Ta có \(A=3+3^2+3^3+...+3^{60}=3\left(1+3+3^2+...+3^{59}\right)⋮3\)
b. Ta có \(A=3+3^2+3^3+...+3^{60}=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{59}\left(1+3\right)=4\left(3+3^3+...+3^{59}\right)⋮4\)c) Ta có \(A=3+3^2+3^3+...+3^{60}=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)=13\left(3+3^4+...+3^{58}\right)⋮13\)
\(A=3+3^2+...+3^{60}\\ =\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\\ =3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...3^{58}\left(1+3+3^2\right)\\ =\left(1+3+3^2\right)\left(3+3^4+...+3^{58}\right)=13\left(3+3^4+...+3^{58}\right)⋮13\)
\(\Rightarrow A⋮13\)
c) chia hÊts cho 13
M= (3\(^3\)+ 3+3\(^2\))+ ...(3\(^{58}+3^{59}+3^{60}\))
M= 3. (1+3+9)+..+ 3\(^{58}\).(1+3+9)
M= 3.13 +....+ 3\(^{58}\). 13
M= 13 .(3+...+\(3^{58}\))
KL :....
a: Ta có: \(A=3+3^2+3^3+...+3^{60}\)
\(=3\left(1+3+3^2+...+3^{59}\right)⋮3\)
b: Ta có: \(A=3+3^2+3^3+3^4+...+3^{60}\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{59}\left(1+3\right)\)
\(=4\cdot\left(3+3^3+...+3^{59}\right)⋮4\)
c: Ta có: \(A=3+3^2+3^3+3^4+...+3^{60}\)
\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\)
\(=13\cdot\left(3+3^4+...+3^{58}\right)⋮13\)