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a)116+115=(..................1)+(..................1)=..........................2
Vì có chữ số tận cùng là 2 nên chia hết cho 4
Bài này thì chắc phải dùng đồng dư -_-
a) Ta có:
11 đồng dư với -1 (mod 4) => 115 đồng dư với (-1)5 = -1 (mod 4) => 115 + 1 chia hết cho 4
=> 116 đồng dư với (-1)6 (mod 4)
=> 116 đồng dư với 1 (mod 4)
=> 116 - 1 chia hết cho 4
=> (116 - 1) + (115 + 1) chia hết cho 4
=> 116 + 115 chia hết cho 4
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+..+\left(2^{59}+2^{60}\right)=3.2+3.2^3+3.2^5+..+3.2^{59}\) 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
a) \(A=2+2^2+...+2^{120}\)
\(\Rightarrow A=\left(2+2^2\right)+...+\left(2^{119}+2^{120}\right)\)
\(\Rightarrow A=\left(2+2^2\right)+...+2^{118}.\left(2+2^2\right)\)
\(\Rightarrow A=6+...+2^{118}.6\)
\(\Rightarrow A=6.\left(1+...+2^{118}\right)⋮3\Rightarrow A⋮3\left(đpcm\right)\)
b) \(A=2+2^2+...+2^{120}\)
\(\Rightarrow A=\left(2+2^2+2^3\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\)
\(\Rightarrow A=\left(2+2^2+2^3\right)+...+2^{117}.\left(2+2^2+2^3\right)\)
\(\Rightarrow A=14+...+2^{117}.14\)
\(\Rightarrow A=14.\left(1+...+2^{117}\right)⋮7\Rightarrow A⋮7\left(đpcm\right)\)
A = 21 + 22 + 23 + ................ + 2120
Chứng minh chia hết cho 7
A = 21 + 22 + 23 + ................ + 2120
A = (21 + 22 + 23) + (24 + 25 + 26) + ................ + (2118 + 2119 + 2120)
A = 2.(1 + 2 + 4) + 24.(1 + 2 + 4) + ................. + 2118.(1 + 2 + 4)
A = 2.7 + 24 . 7 + ................ + 2118.7
A = 7.(2 + 24 + ........... + 2118)
Chứng minh chia hết cho 31
A = 21 + 22 + 23 + ................ + 2120
A = (21 + 22 + 23 + 24 + 25) + (26 + 27 + 28 + 29 + 210) + ................ + (2116 + 2117 + 2118 + 2119 + 2120)
A = 2.(1 + 2 + 4 + 8 + 16) + 26.(1 + 2 +4 + 8 + 16) + ............. + 2116.(1 + 2 + 4 + 8 + 16)
A = 2.31 + 26.31 + ....... + 2116 . 31
A = 31.(2 + 26 + ........... + 2116)
a) \(A=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(A=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\)
\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)
\(A=3\left(2+2^3+...+2^{2009}\right)⋮3\)
\(A=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(A=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)
\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)
\(A=7\left(2^1+2^4+...+2^{2008}\right)⋮7\)
Các ý dưới bạn làm tương tự nhé.
a) \(A=2+2^2+2^3+\dots+2^{60}\)
\(2A=2^2+2^3+2^4+\dots+2^{61}\)
\(2A-A=\left(2^2+2^3+2^4+\dots+2^{61}\right)-\left(2+2^2+2^3+\dots+2^{60}\right)\)
\(A=2^{61}-2\)
Vậy: \(A=2^{61}-2\).
b)
+) \(A=2+2^2+2^3+\dots+2^{60}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\left(2^5+2^6\right)+\dots+\left(2^{59}+2^{60}\right)\)
\(=2\cdot\left(1+2\right)+2^3\cdot\left(1+2\right)+2^5\cdot\left(1+2\right)+\dots+2^{59}\cdot\left(1+2\right)\)
\(=2\cdot3+2^3\cdot3+2^5\cdot3+\dots+2^{59}\cdot3\)
\(=3\cdot\left(2+2^3+2^5+\dots+2^{59}\right)\)
Vì \(3\cdot\left(2+2^3+2^5+\dots+2^{59}\right)⋮3\) nên \(A⋮3\)
+) \(A=2+2^2+2^3+\dots+2^{60}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\left(2^9+2^{10}+2^{11}+2^{12}\right)+\dots+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(=2\cdot\left(1+2+2^2+2^3\right)+2^5\cdot\left(1+2+2^2+2^3\right)+2^9\cdot\left(1+2+2^2+2^3\right)+\dots+2^{57}\cdot\left(1+2+2^2+2^3\right)\)
\(=2\cdot15+2^5\cdot15+2^9\cdot15+\dots+2^{57}\cdot15\)
\(=15\cdot\left(2+2^5+2^9+\dots+2^{57}\right)\)
Vì \(15⋮5\) nên \(15\cdot\left(2+2^5+2^9+\dots+2^{57}\right)⋮5\)
hay \(A\vdots5\)
+) \(A=2+2^2+2^3+\dots+2^{60}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\left(2^7+2^8+2^9\right)+\dots+\left(2^{58}+2^{59}+2^{60}\right)\)
\(=2\cdot\left(1+2+2^2\right)+2^4\cdot\left(1+2+2^2\right)+2^7\cdot\left(1+2+2^2\right)+\dots+2^{58}\cdot\left(1+2+2^2\right)\)
\(=2\cdot7+2^4\cdot7+2^7\cdot7+\dots+2^{58}\cdot7\)
\(=7\cdot\left(2+2^4+2^7+\dots+2^{58}\right)\)
Vì \(7\cdot\left(2+2^4+2^7+\dots+2^{58}\right)⋮7\) nên \(A⋮7\)
$Toru$
a) �=2+22+23+⋯+260A=2+22+23+⋯+260
2�=22+23+24+⋯+2612A=22+23+24+⋯+261
2�−�=(22+23+24+⋯+261)−(2+22+23+⋯+260)2A−A=(22+23+24+⋯+261)−(2+22+23+⋯+260)
�=261−2A=261−2
Vậy: �=261−2A=261−2.
b)
+) �=2+22+23+⋯+260A=2+22+23+⋯+260
=(2+22)+(23+24)+(25+26)+⋯+(259+260)=(2+22)+(23+24)+(25+26)+⋯+(259+260)
=2⋅(1+2)+23⋅(1+2)+25⋅(1+2)+⋯+259⋅(1+2)=2⋅(1+2)+23⋅(1+2)+25⋅(1+2)+⋯+259⋅(1+2)
=2⋅3+23⋅3+25⋅3+⋯+259⋅3=2⋅3+23⋅3+25⋅3+⋯+259⋅3
=3⋅(2+23+25+⋯+259)=3⋅(2+23+25+⋯+259)
Vì 3⋅(2+23+25+⋯+259)⋮33⋅(2+23+25+⋯+259)⋮3 nên �⋮3A⋮3
+) �=2+22+23+⋯+260A=2+22+23+⋯+260
=(2+22+23+24)+(25+26+27+28)+(29+210+211+212)+⋯+(257+258+259+260)=(2+22+23+24)+(25+26+27+28)+(29+210+211+212)+⋯+(257+258+259+260)
=2⋅(1+2+22+23)+25⋅(1+2+22+23)+29⋅(1+2+22+23)+⋯+257⋅(1+2+22+23)=2⋅(1+2+22+23)+25⋅(1+2+22+23)+29⋅(1+2+22+23)+⋯+257⋅(1+2+22+23)
=2⋅15+25⋅15+29⋅15+⋯+257⋅15=2⋅15+25⋅15+29⋅15+⋯+257⋅15
=15⋅(2+25+29+⋯+257)=15⋅(2+25+29+⋯+257)
Vì 15⋮515⋮5 nên 15⋅(2+25+29+⋯+257)⋮515⋅(2+25+29+⋯+257)⋮5
hay �⋮5A⋮5
+) �=2+22+23+⋯+260A=2+22+23+⋯+260
=(2+22+23)+(24+25+26)+(27+28+29)+⋯+(258+259+260)=(2+22+23)+(24+25+26)+(27+28+29)+⋯+(258+259+260)
=2⋅(1+2+22)+24⋅(1+2+22)+27⋅(1+2+22)+⋯+258⋅(1+2+22)=2⋅(1+2+22)+24⋅(1+2+22)+27⋅(1+2+22)+⋯+258⋅(1+2+22)
=2⋅7+24⋅7+27⋅7+⋯+258⋅7=2⋅7+24⋅7+27⋅7+⋯+258⋅7
=7⋅(2+24+27+⋯+258)=7⋅(2+24+27+⋯+258)
Vì 7⋅(2+24+27+⋯+258)⋮77⋅(2+24+27+⋯+258)⋮7 nên �⋮7A⋮7
*Ta có: A\(=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(=\left(2+2^2\right)+2^2\times\left(2+2^2\right)+...+2^{2008}\times\left(2+2^2\right)\)
\(=\left(2+2^2\right)\times\left(1+2^2+2^3+...+2^{2008}\right)\)
\(=6\times\left(2^2+2^3+...+2^{2008}\right)\)
\(=3\times2\times\left(2^2+2^3+...+2^{2008}\right)\)
\(\Rightarrow A⋮3\)
*Ta có: A \(=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(=2\times\left(1+2+2^2\right)+2^4\times\left(1+2+2^2\right)+...+2^{2008}\times\left(1+2+2^2\right)\)
\(=\left(1+2+2^2\right)\times\left(2+2^4+2^7+...+2^{2008}\right)\)
\(=7\times\left(2+2^4+2^7+...+2^{2008}\right)\)
\(\Rightarrow A⋮7\)
Mình sửa lại đề C 1 chút xíu
*Ta có: C \(=3^1+3^2+3^3+3^4+...+3^{2010}\)
\(=\left(3+3^2\right)+3^2\times\left(3+3^2\right)+...+3^{2008}\times\left(3+3^2\right)\)
\(=\left(3+3^2\right)\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(=12\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(=4\times3\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(\Rightarrow C⋮4\)
Các câu khác làm tương tự nhé. Chúc bạn học tốt!
Ta có:
\(A=2+2^2+2^3+2^4+...+2^{59}+2^{60}\\ =2\left(1+2+2^2+2^3+...+2^{58}+2^{59}\right)\\ 2⋮2\Rightarrow2\left(1+2+2^2+2^3+...+2^{58}+2^{59}\right)⋮2\Rightarrow A⋮2\)
\(A=2+2^2+2^3+2^4+...+2^{59}+2^{60}\\ =2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\\ 3\left(2+2^3+...+2^{59}\right)\\ Vì...3⋮3\Rightarrow3\left(2+2^3+...+2^{59}\right)⋮3\\ \Rightarrow A⋮3\)
2;3 nguyên tố cùng nhau nên A chia hết cho 2 và 3 thì A chia hết cho 2 x 3 = 6.
Hay A ⋮ 6 (đpcm).
b.
\(A=2+2^2+2^3+...+2^{58}+2^{59}+2^{60}\\ =2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\\ =7.2+...+7.2^{58}\\ =7\left(2+2^4+2^7+...+2^{58}\right)\\ Vì...7⋮7\Rightarrow7\left(2+2^4+2^7+...+2^{58}\right)⋮7\\ \Rightarrow A⋮7.đpcm\)
A = 2 + 22 + 23 +.....+ 260
A = (2 + 22) + ( 23+ 24)+....+(259+ 260)
A = 6 + 22.( 2+22) + ....+ 258.( 2 + 22)
A = 6 + 22. 6 + .....+ 258. 6
A = 6. ( 1 + 22+ ....+ 258)
vì 6 ⋮ 6 ⇔ 6. ( 1 + 22 +....+258) ⋮ 6 ⇔ A ⋮ 6 (đpcm)
b, A = 2 + 22 + 23 + ....+260
A = (2+22 + 23) + (24+25+26)+.....+(258+259+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)
vì 7 ⋮ 7 ⇔ 7.(2 + 24 +....+258) ⇔ A ⋮ 7 (đpcm)