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\(S=1+2+2^2+2^3+...+2^{59}\)
\(S=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{58}+2^{59}\right)\)
\(S=3+2^2\cdot3+...+2^{58}\cdot3\)
\(S=3\cdot\left(1+2^2+...+2^{58}\right)\)
S chia hết cho 3
_____
\(S=1+2+2^2+...+2^{59}\)
\(S=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+...+\left(2^{57}+2^{58}+2^{59}\right)\)
\(S=7+7\cdot2^3+...+7\cdot2^{57}\)
\(S=7\cdot\left(1+2^3+...+2^{57}\right)\)
S chia hết cho 7
_____
\(S=1+2+2^2+2^3+...+2^{59}\)
\(S=\left(1+2+2^2+2^3\right)+\left(2^4+2^5+2^6+2^7\right)+...+\left(2^{56}+2^{57}+2^{58}+2^{59}\right)\)
\(S=15+2^4\cdot15+...+2^{56}\cdot15\)
\(S=15\cdot\left(1+2^4+...+2^{56}\right)\)
S chia hết cho 15
![](https://rs.olm.vn/images/avt/0.png?1311)
b) A=2+22+23+...+220
A=(2+22)+(23+24)+...+(219+220)
A=3.2+3.23+...+3.219
A=3.(2+23+25+...+219)
⇒A⋮3
phần c) làm tương tự
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a: Ta có: \(A=2+2^2+2^3+...+2^{20}\)
\(=2\left(1+2+2^2+...+2^{19}\right)⋮2\)
b: Ta có: \(A=2+2^2+2^3+...+2^{20}\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{19}\left(1+2\right)\)
\(=3\cdot\left(2+2^3+...+2^{19}\right)⋮3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(2A=2^2+2^3+...+2^{61}\)
=>A=2^61-2
b: \(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=7\left(2+2^4+...+2^{55}+2^{58}\right)\) chia hết cho 7(1)
\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)=3\left(2+2^3+...+2^{59}\right)⋮3\left(2\right)\)
Từ (1), (2) suy ra A chia hết cho 21
![](https://rs.olm.vn/images/avt/0.png?1311)
Đề sai, viết lại thành:
A= 21+22+23+24+...+259+260
Giải:
A=21+22+23+...............+259+260
A=(21+22+23)+...............+(258+259+260)
A=2.(1+2+22)+............+258.(1+2+22)
A=2.7+.......................+258.7
A=(2+24+..............+258).7 ⋮ 7(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
A= (2+22)+(23+24)+...+(259+260)
A=2.(1+2)+23.(1+2)+...+259.(1+2)
A=2.3+23.3+...+259.3
A=3.(2+23+...+259)
Vì 3 chia hết cho 3 => 3.(2+23+...+259) chia hết cho 3
=>A chia hết cho 3
A= (2+22+23)+...+(258+259+260)
A=2.(1+2+22)+...+258.(1+2+22)
A=2.7+...+258.7
A=7.(2+...+258)
Vì 7 chia hết cho 7 =>7.(2+...+258) chia hết cho 7
CHIA HẾT CHO 3 :
A= (2+22)+(23+24)+...+(259+260)
A=2.(1+2)+23.(1+2)+...+259.(1+2)
A=2.3+23.3+...+259.3
A=3.(2+23+...+259)
Vì 3 chia hết cho 3 => 3.(2+23+...+259) chia hết cho 3
=>A chia hết cho 3
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=1+2+2^2+...+2^{41}\)
=>\(2A=2+2^2+2^3+...+2^{42}\)
=>\(2A-A=2^{42}-1\)
=>\(A=2^{42}-1\)
b: \(A=\left(1+2\right)+2^2\left(1+2\right)+...+2^{40}\left(1+2\right)\)
\(=3\left(1+2^2+...+2^{40}\right)⋮3\)
\(A=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...+2^{39}\left(1+2+2^2\right)\)
\(=7\left(1+2^3+...+2^{39}\right)⋮7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=2\left(1+2+2^2+...+2^{58}+2^{59}\right)⋮2\)
\(B=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)
\(=3\left(2+2^3+...+2^{59}\right)⋮3\)
\(B=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=7\left(2+2^4+...+2^{58}\right)⋮7\)
\(B=2\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)
\(=15\left(2+2^5+...+2^{57}\right)⋮15\)
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