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A = 2 + 22 + .... + 260
A = 2(1 + 2) + .... + 259(1 + 2)
A = 2. 3 + .... + 259. 3
=> A chia hết cho 3
Bạn làm tương tự với 7 và 5 :))
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\(A=2+2^2+2^3+...+2^{60}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)
\(=3\left(2+2^3+...+2^{59}\right)\)chia hết cho \(3\).
\(A=2+2^2+2^3+...+2^{60}\)
\(=\left(2+2^2+2^3\right)+\left(2^3+2^4+2^5\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=7\left(2+2^4+...+2^{58}\right)\)chia hết cho \(7\).
\(A=2+2^2+2^3+...+2^{60}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)
\(=15\left(2+2^5+...+2^{57}\right)\)chia hết cho \(15\).
Mà \(\left(15,7\right)=1\)nên \(A\)chia hết cho \(7.15=105\).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2+2^2+...+2^{59}+2^{60}\)
\(A=2\left(1+2\right)+...+2^{59}\left(1+2\right)\)
\(A=2\cdot3+...+2^{59}\cdot3\)
\(A=3\cdot\left(2+...+2^{59}\right)⋮3\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 2 + 22 + 23 + ... + 260 chia hết cho 3
A = ( 2 + 22) + ... + ( 259 + 260 )
A = 2. ( 1 + 2 ) + ... + 259. ( 1 + 2 )
A = 2. 3 + ... + 259 . 3 chia hết cho 3 .
A = 2 + 22 + 23 +... + 260 chia hết cho 7
A = 2.( 1 + 2 + 4 ) + ... + 257 . ( 1 + 2 + 4 )
A = 2.7 + .. + 257 . 7 chia hết cho 7 .
Bạn coi lại phần chứng minh A chia hết cho 105 đi nhé !
Nếu bạn nào thấy đúng , nhớ k cho mình nha !
mk nghĩ là không phải chia hết cho 105 đâu
là chia hết cho 15 thì hợp lí hơn
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+..+\left(2^{59}+2^{60}\right)=3.2+3.2^3+3.2^5+..+3.2^{59}\) Vậy A chia hết cho 3
\(A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+..+\left(2^{58}+2^{59}+2^{60}\right)=7.2+7.2^4+..+7.2^{58}\) Vậy A chia hết cho 7
\(A=\left(2+2^2+2^3+2^4\right)+..+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)=2.15+2^5.15+..+2^{57}.15\) Vậy A chia hết cho 15.
\(B=\left(3+3^3+3^5\right)+..+\left(3^{1987}+3^{1989}+3^{1991}\right)=3.91+3^7.91+..+3^{1986}.91\)
mà 91 chia hết cho 13 nên B chia hết cho 13.
\(B=\left(3+3^3+3^5+3^7\right)+..+\left(3^{1985}+3^{1987}+3^{1989}+3^{1991}\right)=3.820+3^9.820+..+3^{1985}.820\)Mà 820 chia hết cho 41 nên B chia hết cho 41.
D : để ý rằng \(11^k\) đều có đuôi là 1
nên D có đuôi là đuôi của \(1+1+..+1=10\)
Vậy D chia hết cho 5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\\ A=\left(2+1\right)\left(1+2^3+...+2^{59}\right)\\ A=3\left(1+2^3+...+2^{59}\right)⋮3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(A=2+2^2+2^3+...+2^{60}\)
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)
\(A=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{58}\left(2+2^2\right)\)
\(A=\left(2+2^2\right)\left(1+2^2+...+2^{58}\right)\)
\(A=6\left(1+2^2+...+2^{58}\right)⋮3\)(đpcm thứ nhất)
Lại có: \(A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)
\(A=\left(2+2^2+2^3\right)+2^3\left(2+2^2+2^3\right)+...+2^{57}\left(2+2^2+2^3\right)\)
\(A=\left(2+2^2+2^3\right)\left(1+2^3+...+2^{57}\right)\)
\(A=14\left(1+2^3+...+2^{57}\right)⋮7\)(đpcm thứ hai)
Mặt khác: \(A=\left(2+2^3\right)+\left(2^2+2^4\right)+...+\left(2^{58}+2^{60}\right)\)
\(A=\left(2+2^3\right)+2\left(2+2^3\right)+...+2^{57}\left(2+2^3\right)\)
\(A=\left(2+2^3\right)\left(1+2+...+2^{57}\right)\)
\(A=10\left(1+2+...+2^{57}\right)⋮5\)(đpcm thứ ba)