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NV
0
28 tháng 4 2015
A= (21+22+23)+(24+25+26)+...+(258+259+260)
=20(21+22+23)+23(21+22+23)+...+257(21+22+23)
=(21+22+23)(20+23+...+257)
= 14(20+23+...+257) chia hết cho 7
Vậy A chia hết cho 7
25 tháng 6 2015
gọi 1/41+1/42+1/43+...+1/80=S
ta có :
S>1/60+1/60+1/60+...+1/60
S>1/60 x 40
S>8/12>7/12
Vậy S>7/12
CC
2
9 tháng 7 2017
TA có:VÌ 2= 2^1
A=\(2^1+2^2+2^3+...+2^{60}\)
A= \(\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)
A= \(2\left(1+2\right)+2^3\left(1+2\right)+...2^{59}\left(1+2\right)\)
A= \(3.\left(2+2^3+...+2^{60}\right)\)chia hết cho 3
=) A chia hết cho3( đpcm)
Ta lại có:
A= \(2^1+2^2+2^3+...+2^{60}\)
A= \(\left(2^1+2^2+2^3\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)
A=\(2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
A= \(7.\left(2+...+2^{58}\right)\)chia hết cho 7
=) A chia hết cho 7( đpcm)
T
1 tháng 12 2023
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 \(A⋮7\)
$Toru$
1 tháng 12 2023
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
1L
1
CT
1
H
20 tháng 10 2018
b, B= 2 +22 + 23 + 24 + .... + 260
=> B= 2 . 1 + 2 . 2 + 22 . 2 + 23 . 2 + ..... + 259. 2
=> B= 2. ( 1 + 2 + 22 + 23 + ... + 259)
\(\Rightarrow B⋮2\)
B= 2 +22 + 23 + 24 + .... + 260
=> B = ( 2 +22 ) + ( 23 + 24) + .... + ( 259 + 260)
=> B = 2. ( 1 + 2 ) + 23..( 1 + 2 ) + .... + 259. ( 1 + 2 )
=> B = 3 . ( 2 + 23 + ... + 259)
\(\Rightarrow B⋮3\)
B= 2 +22 + 23 + 24 + .... + 260
=> B = ( 2 +22 + 23 ) + ( 24 + 25 + 26 ) + .... ( 258+ 259+ 260)
=> B= 2 . ( 1 + 2 + 22 ) + 24 . ( 1 + 2 + 22 ) + ... + 258. ( 1 + 2 + 22)
B = 7 . ( 2 + 24 + ... + 258)
\(\Rightarrow B⋮7\)
tương tự chia hết cho 15
ghép 4 số và chung là : 1 + 2 + 22 + 23
NM
0
\(2+2^2+2^3+...+2^{60}\\ =\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\\ =2.15+2^5.15+...+2^{57}.15=15\left(2+2^5+...+2^{57}\right)\)
Mà \(15\left(2+2^5+...+2^{57}\right)⋮3\) và \(15\left(2+2^5+...+2^{57}\right)⋮5\) nên A chia hết cho 3 và 5