Đáp án:
Giải thích các bước giải:
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 = \(2\) × \(\left(2+1\right)+2^3\) × \(\left(2+1\right)\) \(+...+2^{59}\) × \(\left(2+1\right)\)
A = \(2\) × \(3+2^3\) × 3 \(+...+2^{59}\) × \(3\)
A = \(3\) × \(\left(2+2^3+...+2^{59}\right)\)
Vì A = \(3\) × \(\left(2+2^3+...+2^{59}\right)\) nên A chia hết cho 3
A = \(\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)\) \(+...+\left(2^{58}+2^{59}+2^{60}\right)\)
A = \(2\) × \(\left(1+2+2^2\right)+2^4\)× \(\left(1+2+2^2\right)+...+2^{58}\) × \(\left(1+2+2^2\right)\)
A = \(2\) × \(7\) \(+2^4\) × \(7+...+2^{58}\) × \(7\)
A = \(7\) × \(\left(2+2^4+...+2^{58}\right)\)
Vì A = \(7\) × \(\left(2+2^4+...+2^{58}\right)\) nên A chia hết cho 7
A = \(\left(2+2^2+2^3+2^4\right)\) \(+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
A = \(2\) × \(\left(1+2+2^2+2^3\right)+2^5\) x \(\left(1+2+2^2+2^3\right)+...+2^{57}\) × \(\left(1+2+2^2+2^3\right)\)
A = \(2\) × \(15+2^5\) × \(15\) \(+...+2^{57}\) × \(15\)A = \(15\) × \(\left(2+2^4+...+2^{58}\right)\)
Vì A = \(15\) × \(\left(2+2^4+...+2^{58}\right)\) nên A chia hết cho 15
