Các câu hỏi này tương tự nhau
Chia hết cho 3 thì ta nhóm 2 số với nhau
\(B=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)
\(B=2\cdot\left(1+2\right)+2^3\cdot\left(1+2\right)+...+2^{59}\cdot\left(1+2\right)\)
\(B=2\cdot3+2^3\cdot3+...+2^{59}\cdot3\)
\(B=3\cdot\left(2+2^3+...+2^{59}\right)⋮3\left(đpcm\right)\)
Tương tự chia hết cho 7 nhóm 3 số, chia hết cho 15 nhóm 4 số
\(B=2+2^2+2^3+...+2^{60}\)
\(=\left(2+2^2\right)+...+\left(2^{59}+2^{60}\right)\)
\(=2\left(1+2\right)+...+2^{59}\left(1+2\right)\)
\(=2.3+...+2^{59}.3 ⋮ 3\)
\(B=2+2^2+...+2^{60}\)
\(=\left(2+2^2+2^3\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=2.7+...+2^{58}.7 ⋮ 7\)
\(B=2+2^2+...+2^{60}\)
\(=\left(2+2^2+2^3+2^4\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)
\(=2.15+...+2^{57}.15 ⋮ 15\)
Dễ thấy B có 60 số hạng nên:
B=(2+2^2)+(2^3+2^4)+...+(2^59+2^60)
=2(1+2)+2^3(1+2)+...+2^59(1+2)
=3(2+2^3+...+2^59) chia hết cho 3
B=(2+2^2+2^3)+...+(2^58+2^59+2^60)
=2(1+2+4)+...+2^58(1+2+4)
=7.(2+...+2^58) chia hết cho 7
B=(2+2^2+2^3+2^4)+...+(2^57+2^58+2^59+2^60)
=2(1+2+4+8)+...+2^57(1+2+4+8)
=15.(2+...+2^57) chia hết cho 15
\(B=2+2^2+2^3+2^4+....+2^{60}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+.....+\left(2^{59}+2^{60}\right)\)
\(=2.\left(1+2\right)+2^3.\left(1+2\right)+.....+2^{59}.\left(1+2\right)\)
\(=3.\left(2+2^3+....+2^{59}\right)⋮3\)
Vậy B chia hết cho 3
\(B=2+2^2+2^3+2^4+2^5+....+2^{60}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+....+\left(2^{58}+2^{59}+2^{60}\right)\)
\(=2.\left(1+2+2^2\right)+2^4.\left(1+2+2^2\right)+....+2^{58}.\left(1+2+2^2\right)\)
\(=7.\left(2+2^4+....+2^{58}\right)⋮7\)
Vậy B chia hết cho 7
\(B=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(=2.\left(1+2+2^2+2^3\right)+2^5.\left(1+2+2^2+2^3\right)+...+2^{57}.\left(1+2+2^2+2^3\right)\)
\(=15.\left(2+2^5+....+2^{57}\right)⋮15\)
Vậy B chia hết cho 15