cho a^2 +ab + b^2 chia hết cho 10. CMR (a^3 - b^3) chia hết cho 1000
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b)=3^1+(3^2+3^3+3^4)+(3^5+3^6+3^7)+....+(3^58+3^59+3^60)
=3^1+(3^2.1+3^2.3+3^2.9)+(3^5.1+3^5.3+3^5.9)+......+(3^58.1+3^58.3+3^58.9)
=3^1+3^2.(1+3+9)+3^5.(1+3+9)+.....+3^58.(1+3+9)
=3+3^2.13+3^5.13+.........+3^58.13
=3.13.(3^2+3^5+....+3^58)
vi tich tren co thua so 13 nen tich do chia het cho 13
=
bai1
a) A=(31+32)+(33+34)+...+(359+360)
=(3^1.1+3^1.3)+...+(3^59.1+3^59.2)
=3^1.(1+3)+...+3^59.(1+3)
=3^1.4+....+3^59.4
=4.(3^1+...+3^59)
vi tich tren co thua so 4 nen tich do chia het cho 4
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+ ) A = 2 + 22 + 23 + 24 + ... + 22003 + 22004
=> A = ( 2 + 22 ) + ( 23 + 24 ) + .... + ( 22003 + 22004 )
=> A = 2.( 1 + 2 ) + 23.( 1 + 2 ) + ... + 22003.( 1 + 2 )
=> A = 2.3 + 23.3 + .... + 22003.3
=> A = 3.( 2 + 23 + 25 + .... + 22001 + 22003 )
Vì 3 ⋮ 3 => A ⋮ 3 ( ĐPCM )
+ ) A = 2 + 22 + 23 + 24 + 25 + 26 + ..... + 22002 + 22003 + 22004
=> A = ( 2 + 22 + 23 ) + ( 24 + 25 + 26 ) + .... + ( 22002.22003.22004 )
=> A = 2.( 1 + 2 + 2.2 ) + 24.( 1 + 2 + 2.2 ) + ... + 22002.( 1 + 2 + 2.2 )
=> A = 2.7 + 24.7 + 27.7 + .... + 22002.7
=> A = 7.( 2 + 24 + 27 + ... + 22002 )
Vì 7 ⋮ 7 => A ⋮ 3 ( ĐPCM )
+ ) A = 2 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + .... + 22001 + 22002 + 22003 + 22004
=> A = ( 2 + 22 + 23 + 24 ) + ( 25 + 26 + 27 + 28 ) + .... + ( 22001 + 22002 + 22003 + 22004 )
=> A = 2.( 1 + 2 + 22 + 23 ) + 25.( 1 + 2 + 22 + 23 ) + .... + 22001.( 1 + 2 + 22 + 23 )
=> A = 2.15 + 25.15 + 29.15 + .... + 22001.15
=> A = 15.( 2 + 25 + 29 + .... + 22001 )
Vì 15 ⋮ 15 => A ⋮ 15 ( ĐPCM )
Câu b tương tự .
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\(\text{a) }a+b\text{ chia hết cho 3}\)
\(\Rightarrow a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\) chia hết cho 3
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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