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a) B\(=\) 3 + 32 + 33 + ... + 360
\(=\)(3+32)+(33+34)+...+(359+360)
\(=\)3(1+3)+33(1+3)+...+359(1+3)
\(=\)(3+1)(3+33+...+359)
\(=\)4(3+33+...+359)⋮4
⇒B⋮4
b) B\(=\)(3+32+33)+...+(358+359+360)
\(=\)30(3+32+33)+...+357(358+359+360)
\(=\)3+32+33(30+33+36+...+357)
\(=\)39(30+33+36+...+357)⋮13
⇒ B⋮13
\(B=3+3^2+3^3+....+3^{120}\)
a, Ta thấy : Cách số hạng của B đều chi hết cho 3
\(B=3+3^2+3^3+....+3^{120}⋮3\)
\(b,B=3+3^2+3^3+....+3^{120}\)
\(B=\left(3+3^2\right)+\left(3^3+3^4\right)+....+\left(3^{119}+3^{120}\right)\)
\(B=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{119}\left(1+3\right)\)
\(B=3.4+3^3.4+...+3^{119}.4\)
\(B=4\left(3+3^3+...+3^{199}\right)\)
Có : \(B=4\left(3+3^3+...+3^{199}\right)⋮4\)
\(\Rightarrow B⋮4\)
\(c,B=3+3^2+3^3+....+3^{120}\)
\(B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{119}+3^{120}\right)\)
\(B=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{118}\left(3+3^2\right)\)
\(B=13+3^2.13+...+3^{118}.13\)
\(B=13\left(3^2+3^4+...+3^{118}\right)\)
Có : \(B=13\left(3^2+3^4+...+3^{118}\right)⋮13\)
\(\Rightarrow B⋮13\)
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
a: \(B=3\left(1+3+3^2+...+3^{120}\right)⋮3\)
b: \(B=4\left(3+...+3^{119}\right)⋮4\)
a/
\(A=3\left(1+3+3^2\right)+...+3^{118}\left(1+3+3^2\right)=\)
\(=13\left(3+3^4+3^7+...+3^{118}\right)⋮13\)
\(A=3\left(1+3+3^2+3^3\right)+...+3^{117}\left(1+3+3^2+3^3\right)=\)
\(A=40\left(3+3^5+3^9+...+3^{117}\right)⋮40\)
b/
\(A=3+3^2\left(1+3+3^2+...+3^{118}\right)=\)
\(=3+9\left(1+3+3^2+...+3^{118}\right)\) chia 9 dư 3 nên A không chia hết cho 9
c/
\(3A=3^2+3^3+3^4+...+3^{121}\)
\(\Rightarrow2A=3A-A=3^{121}-3\Rightarrow2A+3=3^{121}\)
\(2A+3=3^{121}=3.3^{120}=3.\left(3^4\right)^{30}=3.81^{30}\) có tận cùng là 3 nên 2A+3 không phải là số chính phương
\(B=3+3^2+3^3+...+3^{118}+3^{119}+3^{120}\\ =\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{118}+3^{119}+3^{120}\right)\\ =3.\left(1+3+3^2\right)+3^4.\left(1+3+3^2\right)+...+3^{118}\left(1+3+3^2\right)\\ =\left(3+3^4+...+3^{118}\right).\left(1+3+3^2\right)\\ =\left(3+3^4+...+3^{118}\right).13⋮13\left(ĐPCM\right)\)
a) \(B=3+3^2+3^3+...+3^{120}\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{199}\left(1+3\right)\)
\(=3.4+3^3.4+3^{199}.4=4\left(3+3^3+...+3^{199}\right)⋮4\)
b) \(B=3+3^2+3^3+...+3^{120}\)
\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{198}\left(1+3+3^2\right)\)
\(=3.13+3^4.13+...+3^{198}.13=13\left(3+3^4+...+3^{198}\right)⋮13\)