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Bài 1:
a) +) \(A=2+2^2+...+2^{2004}\)
\(\Rightarrow A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2003}+2^{2004}\right)\)
\(\Rightarrow A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2003}\left(1+2\right)\)
\(\Rightarrow A=2.3+2^3.3+...+2^{2003}.3\)
\(\Rightarrow A=\left(2+2^3+...+2^{2003}\right).3⋮3\)
\(\Rightarrow A⋮3\left(đpcm\right)\)
+) \(A=2+2^2+...+2^{2004}\)
\(\Rightarrow A=\left(2+2^2+2^3\right)+...+\left(2^{2002}+2^{2003}+2^{2004}\right)\)
\(\Rightarrow A=2\left(1+2+2^2\right)+...+2^{2002}\left(1+2+2^2\right)\)
\(\Rightarrow A=2.7+...+2^{2002}.7\)
\(\Rightarrow A=\left(2+...+2^{2002}\right).7⋮7\)
\(\Rightarrow A⋮7\left(đpcm\right)\)
+) \(A=2+2^2+....+2^{2004}\)
\(\Rightarrow A=\left(2+2^2+2^3+2^4\right)+...+\left(2^{2001}+2^{2002}+2^{2003}+2^{2004}\right)\)
\(\Rightarrow A=2\left(1+2+2^2+2^3\right)+...+2^{2001}\left(1+2+2^2+2^3\right)\)
\(\Rightarrow A=2.15+...+2^{2001}.15\)
\(\Rightarrow A=\left(2+...+2^{2001}\right).15⋮15\)
\(\Rightarrow A⋮15\left(đpcm\right)\)
b) \(B=1+3+3^2+...+3^{99}\)
\(\Rightarrow B=\left(1+3+3^2+3^3\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\)
\(\Rightarrow B=\left(1+3+9+27\right)+...+3^{96}\left(1+3+3^2+3^3\right)\)
\(\Rightarrow B=40+...+3^{96}.40\)
\(\Rightarrow B=\left(1+...+3^{96}\right).40⋮40\)
\(\Rightarrow B⋮40\left(đpcm\right)\)
1,Chứng minh chia hết cho 3
A=2+2^2+2^3+2^4+2^5+2^6+2^7+...+2^2004
A=(2+2^2)+(2^3+2^4)+(2^5+2^6)+...+(2^2003+2^2004)
A=2(1+2)+2^3(1+2)+2^5(1+2)+...+2^2003(1+2)
A=2.3+2^3.3+2^5.3+..+2^2003.3
A=(2+2^3+2^5+...+2^2003).3 chia hết cho 3 (đpcm)
chứng minh chia hết cho 7
A=(2+2^2+2^3)+(2^4+2^5+2^6)+...+(2^2002+2^2003+2^2004)
A=2(1+2+2^2)+2^4(1+2+2^2)+...+2^2002(1+2+2^2)
A=2.7+2^4.7+...+2^2002.7
A=(2+2^4+..+2^2002).7 chia hết cho 7 (Đpcm)<mik sẽ làm tiếp>
a) A = 2 + 2^2 + ... + 2^58 + 2^59 + 2^60
A = 2 ( 2 + 1 ) + 2^3 ( 2 + 1 ) + ... + 2^59 ( 2 + 1)
A = 3 .2 + 3.2^3 + ... + 3.2^59
A = 3 ( 2 + 2^3 + ... + 2^59 ) luôn chia hết cho 3
Ta có A = 2+22 + 23 + .....+ 259 + 260
= ( 2+ 22 + 23) +....+ (258 + 259 + 260)
= 2(1+2+4) +....+ 258( 1+2+4)
= 2 .7+24.7 +....+ 258 . 7
= 7( 2+24 + ....+ 258)
=> A chia hết cho 7
2004 chia hết cho 3 và cho 4 nên ta có thể lập tổ hợp sau:
\(A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2002}+2^{2003}+2^{2004}\right)\)
\(A=2\cdot\left(1+2+4\right)+2^4\cdot\left(1+2+4\right)+...+2^{2002}\cdot\left(1+2+4\right)=7\cdot\left(2+2^4+...+2^{2002}\right)\)
=> A chia hết cho 7. (1)
Mặt khác:
\(A=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{2001}+2^{2002}+2^{2003}+2^{2004}\right)\)
\(A=2\cdot\left(15\right)+2^5\cdot\left(15\right)+...+2^{2001}\cdot\left(15\right)=15\cdot\left(2+2^5+...+2^{2001}\right)\)=> Achia hết cho 15 (2)
A chia hết cho 15 có nghĩa là A cũng chia hết cho 3 (3).
Từ (1) (2) (3) suy ra ĐPCM.
b) Ta có: A = 2 + 22 + 23 + ....+ 22004
=> A = ( 2 + 22) + ( 23 + 24 ) +....+( 22003 + 22004 )
=> A = 2 x ( 1 + 2) + 23 x ( 1+2) +.....+ 22003 x ( 1 + 2)
=> A = 2 x 3 + 23 x 3 + ......+ 22003 x 3
=> A = 3 x ( 2 + 23 + ...+ 22003 ) chia hết cho 3
=> A chia hết cho 3 (1)
Tương tự: A = 2 + 22 + 23 + 24 + ....+ 22004
= > A = ( 2 + 23 + 23) + ( 24 + 25 + 26) +.....+( 22002 + 22003 + 22004 )
=> A = 2 x ( 1 + 2 + 22) + 24 x ( 1 + 2 + 22 ) +....+ 22002 x ( 1 + 2 + 22 )
=> A = 2 x 7 + 24 x 7 +.........+ 22002 x 7
=> A = 7 x ( 2 + 24 + ....+ 22002 ) chia hết cho 7
=> A chia hết cho 7 (2)
Ta lại có: A = 2 + 22 + 23 + ....+22004
=> A = ( 2 + 22 + 23 + 24 ) + ( 25 + 26 + 27 + 28 ) +....+ ( 22001 + 22002 + 22003 + 22004 )
=> A = 2 x ( 1 + 2 + 22 + 23 ) + 25 x ( 1 + 2 + 22 + 23) +....+ 22001 x ( 1 + 2 + 22 + 23 )
=> A = 2 x 15 + 25 x 15 + ....+ 22001 x 15
=> A = 15 x ( 2 + 25 +....+ 22001 ) chia hết cho 15
=> A chia hết cho 15 (3)
Từ (1);(2);(3) => A chia hết cho 3 ; 7 ; 15
CHÚC BẠN HỌC TỐT
a 2+22=2.1+2.2=2.[1+2]=2.3
2+22+23=2.1+2.2+22.2=2.[1+2+22]
2+22+23+24=2.1+2.2+22.2+23.2=2.[1+2+22+23]
b [2.1+2.2]+[23.1+23.2]+...+[22003.1+22003.2]
=[2[1+2]]+[23[1+2]]+...+[22003[1+2]]
=>A chia het cho 3
cac phep con lai tuong tu