Tổng sau có chia hết cho 3 không?
A=2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^10
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A=(2+2^2)+(2^3+2^4)+(2^5+2^6)+(2^7+2^8)+(2^9+2^10)
A=(2.1+2.2)+...+(2^9.1+2^9.2)
A=2.3+2^3.3+...+2^9.3
A=3.(2+2^3+...+2^9) chia hết cho 3
=> A chia hết cho 3
\(A=2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^{10}.\)
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+......+\left(2^9+2^{10}\right)\)
\(A=2.\left(1+2\right)+2^3.\left(1+2\right)+......2^9\left(1+2\right)\)
\(A=2.3+2^3.3+.......+2^9.3\)
\(A=3.\left(2+2^3+....+2^9\right)\)
Vậy \(A⋮3\)
có bạn cộng 2+ 2^2 rồi gộp các số tiếp theo như thế sẽ biết
A= (2+2^2)+(2^3+2^4)+(2^5+2^6)+(2^7+2^8)+(2^9+2^10)
=2.3+2^3.3+2^5.3+2^7.3+2^9.3=3(2+2^3+2^5+2^7+2^9) chia hết cho 3
A= 2+(2^2+2^3+2^4)+(2^5+2^6+2^7)+(2^8+2^9+2^10)
=2+2^2(1+2+2^2)+2^5(1+2+2^2)+2^8(1+2+2^2)=2+7(2^2+2^5+2^8) không chia hết cho 7
=> A không chia hết cho 21
A = ( 2 + 22 ) + ( 23 + 24 ) + ( 25 + 26 ) + ( 27 + 28 ) + ( 29 + 210 )
A = 2 . 3 + 23 . 3 + 25 . 3 + 27 . 3 + 29 . 3
A = 3 . ( 2 + 23 + 25 + 27 + 29 ) \(⋮\) 3
A = 2 + ( 22 + 23 +24 ) + ( 25 + 26 + 27 ) + ( 28 + 29 + 210 )
A = 2 + 22 . ( 1 + 2 + 22 ) + 25 . ( 1 + 2 + 22 ) + 28 . ( 1 + 2 + 22 )
A = 2 + 7 . ( 22 + 25 + 28 ) \(⋮̸\) 7
=> A \(⋮̸\) 21
hok tốt
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\)
\(A=2.\left(1+2\right)+2^3.\left(1+2\right)+...+2^9.\left(1+2\right)\)
\(A=2.3+2^3.3+...+2^9.3\)
\(A=3.\left(2+2^3+...+2^9\right)⋮3\)
A = 2 + 22 +...+ 210 ( có 10 số hạng)
A = (2+22 ) +( 23+24) + ...+ (29+210)
A = 2.(1+2) + 23.(1+2) + ...+ 29.(1+2)
A = 2.3 + 23.3 + ...+ 29.3
A = 3.(2+23 +...+29) chia hết cho 3
Tổng A có: (10-1):1+1=10(số). Ta nhóm như sau:
A=(2+2^2)+(2^3+2^4)+...+(2^9+2^10)
A=2(1+2)+2^3(1+2)+...+2^9(1+2)
A=2.3+2^3.3+...+2^9.3
A=3(2+2^3+...+2^9) chia hết cho 3
\(A=2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^{10}\)
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+\left(2^5+2^6\right)+\left(2^7+2^8\right)+\left(2^9+2^{10}\right)\)
\(A=2\left(1+2\right)+2^3\left(1+2\right)+2^5\left(1+2\right)+2^7\left(1+2\right)+2^9\left(1+2\right)\)
\(A=2\cdot3+2^3\cdot3+2^5\cdot3+2^7\cdot3+2^9\cdot3\text{ }⋮\text{ }3\)
\(A=2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^{10}\)
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+\left(2^5+2^6\right)+\left(2^7+2^8\right)+\left(2^9+2^{10}\right)\)
\(A=2\left(1+2\right)+2^3\left(1+2\right)+2^5\left(1+2\right)+2^7\left(1+2\right)+2^9\left(1+2\right)\)
\(A=2\cdot3+2^3\cdot3+2^5\cdot3+2^7\cdot3+2^9\cdot3\)
\(\Rightarrow\text{ }A\text{ }⋮\text{ }3\text{ }\left(\text{ ĐPCM}\right)\)
tổng sau có chia hết cho 3 không?
A= 2+ 2^2+ 2^3+ 2^4+ 2^5+ 2^6+ 2^7+ 2^8+ 2^9+ 2^10
giải chi tiết nha
A= (2+22)+(23+24)+...+(29+210)
A= 2.(1+2)+23.(1+2)+...29.(1+2)
A= 2.3+23.3+...+29.3
A= 3.(2+23+25+27+29)
Vậy a chia hết cho 3
=> A = ( 2+2^2) + (2^3+2^4) +....+ (2^9+2^10)
=> A =2(1+2) + 2^3( 1+2)+.....+2^9(1+2)
=> A = 2.3+2^3.3+....+2^9.3
=>A =(2+2^3+....+2^9) .3 Luôn chia hết cho 3
Vậy tổng trên chia hết cho 3
\(A=2+2^2+2^3+...+2^{10}\)
\(\Rightarrow2A=2^2+2^3+2^4+...+2^{11}\)
\(2A-A=A=2^{11}-2=2.\left(2^{10}-1\right)=2.1023=2.3.341\)
Có thừa số 3 nên A chia hết cho 3.