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a) A = \(\left(2+2^2+2^3+...+2^5\right)+\left(2^6+2^7+...+2^{10}\right)\)
\(=\left(2.31\right)+2^5.31=31.\left(2+2^5\right)\)
Vậy A chia hết cho 31
a, A = 2+22+23+...+210
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 (đpcm)
b, A = 2+22+23+...+210
A = (2+22+23+24+25)+(26+27+28+29+210)
A = 2(1+2+22+23+24) + 26.(1+2+22+23+24)
A = 2.31 + 26.31
A = 31.(2+26) chia hết cho 31 (Đpcm)
`B = 2 + 2^3 + 2^5 + 2^7 + ... + 2^31`.
`<=> (2 + 8) + 2^4(2 + 8) + 2^8(2 + 8) + ... + 2^28(2 + 8)`.
`<=> (1 + 2^4 + 2^8 + ... + 2^28)(2+8)`
`<=> 10 . (1 + 2^4 + 2^8 + ... + 2^28)`.
Vì `(1 + 2^4 + ... + 2^28) in ZZ`.
`=> 10 . (1+2^4 + ... + 2^28) vdots 10`.
B=2(1+2^2)+2^5(1+2^2)+...+2^29(1+2^2)
=5(2+2^5+...+2^29)
=10(1+2^4+...+2^28) chia hết cho 10
a) \(\left(1+2+2^2+...+2^7\right)\)
\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^6+2^7\right)\)
\(=\left(1+2\right)+2^2.\left(1+2\right)+...+2^6.\left(1+2\right)\)
\(=3+2^2.3+...+2^6.3\)
\(=3.\left(1+2^2+...+2^6\right)⋮3\left(đpcm\right)\)
a) Đặt A = 1 + 2 + 22 + 23 + ... + 27
Ta có:
A = 1 + 2 + 22 + 23 + ... + 27
\(\Rightarrow\)2A = 2 + 22 + 23 + 24 + ... + 28
\(\Rightarrow\)A = 28 - 1 = 255
Vì 255\(⋮\)3\(\Rightarrow\)2 + 22 + 23 + 24 + ... + 28\(⋮\)3
\(\Rightarrow\)ĐPCM
a=2+2^2+2^3+...+2^10
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=3.(2+2^3+...+2^9)
=> a chia hết cho 3
a=2+2^2+2^3+...+2^10
a=(2+2^2+2^3+2^4+2^5)+(2^6+2^7+2^8+2^9+2^10)
a=2.(1+2+4+8+16)+2^6.(1+2+4+8+16)
a=31.(2+2^6)
=> a chia hết cho 31
chúc bạn học tốt nha
2/
A=1+2+2^2+...+2^10
2.A= 2+2^2+...+2^11
=>2A-A = 2^11-1=> A = 2^11 -1=B
Vậy A=B
1)52003+52002+52001=52001(52+5+1)=52001(25+5+1)=52001.31
Vì 31 chia hết cho 31nên
52001.31chia hết cho 31 hay 52003+52002+52001 chia hết cho 31
2) A = 1+2+22+......+29+210
=>2A=2+22+23+...+211
=>2A-A=2+22+23+...+211-(1+2+22+...+29+210)
=>A=211-1
Vậy A=B=211-1
`#3107`
b)
`B = 2 + 2^2 + 2^3 + ... + 2^9 + 2^10`
`= (2 + 2^2 + 2^3 + 2^4 + 2^5) + (2^6 + 2^7 + 2^8 + 2^9 + 2^10)`
`= 2.(1 + 2 + 2^2 + 2^3 + 2^4) + 2^6 . (1 + 2 + 2^2 + 2^3 + 2^4)`
`= 2.(1 + 2 + 4 + 8 + 16) + 2^6 . (1 + 2 + 4 + 8 + 16)`
`= 2.31 + 2^6.31`
`= 31.(2 + 2^6)`
Vì `31 \vdots 31`
`\Rightarrow 31.(2 + 2^6) \vdots 31`
Vậy, `B \vdots 31 (đpcm).`
B = 2 + 22 + 23 + ... + 29 + 210
B = 2(1 + 21 + ... + 24) + 25(1 + 21 + ... + 24)
B = (2 + 25).(1 + 2 + 4 + 8 + 16)
B = (2 + 25).31
Do đó B chia hết cho 31