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Cái này số nhỏ nên tớ tính luôn nhé :)
A=2+2^2+2^3+......+2^8
=> 2A=2^2+2^3+.......+2^9
=> 2A-A=A=2^9-2=512-2=510 chia hết cho 3(đpcm)
Đặt \(A=2+2^2+2^3+2^4+2^5+2^6+2^7+2^8\)
\(\Rightarrow2A=2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9\)
\(\Rightarrow2A-A=\left(2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9\right)-\left(2+2^2+2^3+2^4+2^5+2^6+2^7+2^8\right)\)
\(\Rightarrow A=2^9-2\)
\(\Rightarrow A=512-2=510⋮3\)
Vậy A chia hết cho 3 (đpcm)
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Ta có: A= 3 + 32 +33 + 34 + ...+ 3100
=> 3A = 32 +33 + 34 + ...+ 3101
=> 3A - A = 3101 - 3
=> 2A = 3101 - 3
=> A = 3101 - 3 / 2
Chứng minh
A = 2 + 22 + 23 + 24 + ...+ 260 chia hết cho 31
A = (2 + 22 + 23 + 24 + 25) + ..... + (256 + 257 + 258 + 259 + 260)
= 2.( 1 + 2 + 4 + 8 + 16) + ..... + 256.(1 + 2 + 4 + 8 + 16)
= 2.31 + .... + 256.31
= 31.(2 + .... + 256) chia hết cho 3`1
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s=2+2^2+2^3+.....+2^100
s=2.(1+2+2^2+2^3)+......+2^97.(1+2+2^2+2^3)
s=2.15+....+2^97.15
s=15.(2+....+2^97)
=> s chia het cho 15
a=3+3^2+3^3+....+3^20
a=3.(1+3)+......+3^19.(1+3)
a=3.4+.....+3^19.4
a=4.(3+.....+3^19)
vay a chia het cho 4
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ta có : \(A=2+2^2+2^3+2^4+...+2^{100}\)
\(\Leftrightarrow A=\left(2+2^2\right)+\left(2^3+2^4\right)+\left(2^5+2^6\right)+...+\left(2^{99}+2^{100}\right)\)
\(\Leftrightarrow A=2\left(1+2\right)+2^3\left(1+2\right)+2^5\left(1+2\right)+...+2^{99}\left(1+2\right)\)
\(\Leftrightarrow A=3\left(2+2^3+2^5+...+2^{99}\right)⋮3\)
\(\Rightarrow\) \(A\) chia hết cho \(3\) (đpcm)
\(A=2+2^2+2^3+2^4+...+2^{100}\)
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\)
\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)\)
\(A=2\cdot3+2^3\cdot3+...+2^{99}\cdot3\)
\(A=3\left(2+2^3+...+2^{99}\right)\)
\(\Rightarrow A⋮3\)
Vậy A\(⋮\)3 (đpcm)
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minh chi lam dc cau a thoi nha nhung hay t i c k cho minh
3 + 32 = 12 chia het cho 4 3 + 32 + 33 + .......+39 + 310 = 30 .[ 3+32 ] + 32 . [ 3 + 32 ] + ....+38 . [ 3 + 32 ]
=30 . 12 + 32 . 12 +.....+ 38 . 12 = 12.[30 + 32 +....+ 38 ]
vi 12 chia het cho 4 nen 12 nhan voi so tu nhien nao thi so do cung chia het cho 4 nen A chia het cho 4