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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)
Lời giải:
$A=(2+2^2+2^3)+(2^4+2^5+2^6)+....+(2^{58}+2^{59}+2^{60})$
$=2(1+2+2^2)+2^4(1+2+2^2)+....+2^{58}(1+2+2^2)$
$=(1+2+2^2)(2+2^4+....+2^{58})$
$=7(2+2^4+....+2^{58})\vdots 7$.
A = 2+22+23+...+260
A = 2.(1+2+22) + 24.(1+2+22) + ... + 258.(1+2+22)
A = 2.7+24.7+...+258.7
A= 7. (2+24+...+258) chia hết cho 7
--> A chia hết cho 7 (ĐPCM)
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a: \(G=8^8+2^{20}\)
\(=2^{24}+2^{20}\)
\(=2^{20}\left(2^4+1\right)=2^{20}\cdot17⋮17\)
b: Sửa đề: \(H=2+2^2+2^3+...+2^{60}\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)
\(=3\left(2+2^3+...+2^{59}\right)⋮3\)
\(H=2+2^2+2^3+...+2^{60}\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=7\left(2+2^4+...+2^{58}\right)⋮7\)
\(H=2+2^2+2^3+...+2^{60}\)
\(=\left(2+2^2+2^3+2^4\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)
\(=15\left(2+2^5+...+2^{57}\right)⋮15\)
c: \(E=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{1989}\left(1+3+3^2\right)\)
\(=13\left(1+3^3+...+3^{1989}\right)⋮13\)
\(E=1+3+3^2+3^3+...+3^{1991}\)
\(=\left(1+3+3^2+3^3+3^4+3^5\right)+\left(3^6+3^7+3^8+3^9+3^{10}+3^{11}\right)+...+3^{1986}+3^{1987}+3^{1988}+3^{1989}+3^{1990}+3^{1991}\)
\(=364\left(1+3^6+...+3^{1986}\right)⋮14\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)
\(=15\left(2+...+2^{57}\right)⋮15\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa dùm mình dòng cuối cùng là " Vậy \(A⋮5\) " nha. Cảm ơn bạn.
Ta có:
\(H=2+2^2+2^3+...+2^{60}\)
\(H=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)
\(H=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\cdot\left(1+2\right)\)
\(H=3\cdot\left(2+2^3+...+2^{59}\right)\)
Vậy H chia hết cho 3
_______
\(H=2+2^2+2^3+...+2^{60}\)
\(H=\left(2+2^2+2^3\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)
\(H=2\cdot\left(1+2+4\right)+2^4\cdot\left(1+2+4\right)+...+2^{58}\cdot\left(1+2+4\right)\)
\(H=7\cdot\left(2+2^4+...+2^{58}\right)\)
Vậy H chia hết cho 7
__________
\(H=2+2^2+2^3+...+2^{60}\)
\(H=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(H=2\cdot\left(1+2+4+8\right)+2^5\cdot\left(1+2+4+8\right)+...+2^{57}\cdot\left(1+2+4+8\right)\)
\(H=15\cdot\left(2+2^5+...+2^{57}\right)\)
Vậy H chia hết cho 15
H=2+22+23+...+260�=2+22+23+...+260
Ta có:
H=2.(1+2)+23.(1+2)+...+259.(1+2)�=2.1+2+23.1+2+...+259.(1+2)
H=2.3+23.3+...+259.3�=2.3+23.3+...+259.3
H=3.(2+23+...+259)⋮3�=3.2+23+...+259 ⋮3
Ta có:
H=2.(1+2+22)+24.(1+2+22)+...+228.(1+2+22)�=2.1+2+22+24.1+2+22+...+228.1+2+22
H=2.7+24.7+...+258.7�=2.7+24.7+...+258.7H=7.(2+24+...+258)⋮7�=7.2+24+...+258 ⋮7
Ta có:
H=2.(1+2+22+23)+25.(1+2+22+23)+...+257.(1+2+22+23)�=2.1+2+22+23+25.1+2+22+23+...+257.1+2+22+23
H=2.15+25.15+...+257.15�=2.15+25.15+...+257.15
H=15.(2+25+...+257)⋮15�=15.2+25+...+257 ⋮15Vậy H chia hết cho 3;7;153; 7; 15.
nhớ tik đúng nha!!!