\(A=2+2^2+2^3+...+2^{60}\\ =\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\\ =2\cdot\left(1+2\right)+2^3\cdot\left(1+2\right)+...+2^{59}\cdot\left(1+2\right)\\ =\left(1+2\right)\cdot\left(2+2^3+...+2^{59}\right)\\ =3\cdot\left(2+2^3+...+2^{59}\right)⋮3\)
\(A=2+2^2+2^3+...+2^{60}\\ =\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\\ =2\cdot\left(1+2+2^2\right)+2^4\cdot\left(1+2+2^2\right)+...+2^{58}\cdot\left(1+2+2^2\right)\\ =\left(1+2+2^2\right)\cdot\left(2+2^4+...+2^{58}\right)\\ =7\cdot\left(2+2^4+...+2^{58}\right)⋮7\)
\(A=2+2^2+2^3+...+2^{60}\\ =\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\\ =2\cdot\left(1+2+2^2+2^3\right)+2^5\cdot\left(1+2+2^2+2^3\right)+...+2^{57}\cdot\left(1+2+2^2+2^3\right)\\ =\left(1+2+2^2+2^3\right)\cdot\left(2+2^5+...+2^{57}\right)\\ =15\cdot\left(2+2^5+...+2^{57}\right)⋮15\)
+A=\(2+2^2+2^3+...+2^{60}\)
+A=\(\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)
+A=\(2.\left(1+2\right)+2^3.\left(1+2\right)+..+2^{59}.\left(1+ 2\right)\)
+A=\(2.3+2^3.3+..+2^{^{ }59}+3\)
=>A chia hết cho 3
Mấy câu sau thì nhóm 3,4 là Ok.
Mình nghĩ là làm như vậy, các bạn thấy thế nào?
