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S = (21+22)+(23+24)+...+(299+2100)
S = 2.(1+2)+23.(1+2)+...+299.(1+2)
S = 2.3+23.3+...+299.3
S = 3.(2+23+...+299)
=> S chia hết cho 3
S = (21+22+23+24)+(25+26+27+28)+...+(297+298+299+2100)
S = 2.(1+2+4+16)+25.(1+2+4+16)+...+297.(1+2+4+16)
S = 2.15+25.15+...+297.15
S = 15.(2+25+...+297)
=> S chia hết cho 15
\(S=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)
\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{88}\left(1+3+3^2\right)\)
\(=3.13+3^4.13+...+3^{88}.13\)
\(=13\left(3+3^4+...+3^{88}\right)\) chia hết cho \(13\)
\(S=3+3^2+3^3+3^4+....+3^{89}+3^{90}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)
\(==3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^{88}\left(1+3+3^2\right)\)
\(=\left(1+3+3^2\right).\left(3+3^4+....+3^{88}\right)\)
\(=13\left(3+3^4+...+3^{88}\right)\)\(⋮\)\(13\)
Ta có :
A = 2 + 22 + ... + 22010
A = ( 2 + 22 ) + ( 23 + 24 ) + ... + ( 22009 + 22010 )
A = 2 . ( 1 + 2 ) + 23 . ( 1 + 2 ) + ... + 22009 . ( 1 + 2 )
A = 2 . 3 + 23 . 3 + ... + 22009 . 3
A = 3 . ( 2 + 23 + ... + 22009 ) \(⋮\)3
A = 2 + 22 + ... + 22010
A = ( 2 + 22 + 23 ) + ( 24 + 25 + 26 ) + ... + ( 22008 + 22009 + 22010 )
A = 2 . ( 1 + 2 + 22 ) + 24 . ( 1 + 2 + 22 ) + ... + 22008 . ( 1 + 2 + 22 )
A = 2 . 7 + 24 . 7 + ... + 22008 . 7
A = 7 . ( 2+ 24 + ... + 22008 ) \(⋮\)7
B = 3 + 32 + ... + 32010
B = ( 3 + 32 ) + ... + ( 32009 + 32010 )
Làm tương tự chứng minh được B \(⋮\)4
B = 3 + 32 + ... + 32010
B = ( 3 + 32 + 33 ) + ... + ( 32008 + 32009 + 32010 )
Làm tương tự chứng minh được B \(⋮\)13
a, \(A=2+2^2+...+2^{2010}\)
\(\Leftrightarrow A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\)
\(\Leftrightarrow A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)\)
\(\Leftrightarrow A=2.3+2^3.3+...+2^{99}.3\)
\(\Leftrightarrow A=3\left(2+2^2+...+2^{99}\right)\)chia hết cho 3
\(B=1+2+2^2+2^3+...+2^{14}+2^{15}\)
\(=1+\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+....+\left(2^{13}+2^{14}+2^{15}\right)\)
\(=1+2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{13}\left(1+2+2^2\right)\)
\(=1+\left(1+2+2^2\right)\left(2+2^4+....+2^{13}\right)\)
\(=1+7\left(2+2^4+...+2^{13}\right)\)
=> B không chia hết cho 7
\(Q=1+3+3^2+3^3+...+3^{19}+3^{20}\)
\(=1+\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{19}+3^{20}\right)\)
\(=1+3\left(1+3\right)+3^3\left(1+3\right)+...+3^{19}\left(1+3\right)\)
\(=1+\left(1+3\right)\left(3+3^3+...+3^{19}\right)\)
\(=1+4\left(3+3^3+...+3^{19}\right)\)
=> Q không chia hết cho 4
a) P=2+22+23+24+...+260 \(⋮\) 21 và 15
\(\Rightarrow\)P = 22+23+24+25+...+261
\(\Rightarrow\) (2P - P) = 261 - 2
\(\Rightarrow\) P = 261 - 2 = 2.(260 - 1)
Để P \(⋮\) 21 và 15 thì (260 - 1) \(⋮\)21 và 15
tức là (260 - 1) \(⋮\)3; 5; 7
*Ta có 260 - 1 = (24)15 = 1615 - 1
= (16 - 1).(1+16+162+163+...+1614)
= 15.(1+16+162+163+...+1614) \(⋮\) 15
Vậy P \(⋮\) 15 (1)
* Ta có 260 - 1 = (26)10 - 1 = 6410 - 1
= (64 - 1).(1+64+642+643+...+649 )
= 63 \(⋮\) (1+64+642+643+...+649 )
= 21.3.(1+64+642+643+...+649 ) \(⋮\) 21
P \(⋮\)21 (2)
Từ (1) và (2) \(\Rightarrow\) P \(⋮\)15 và 21
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