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\(A=1+3+3^2+..........+3^{11}\)
\(\Leftrightarrow A=\left(1+3\right)+\left(3^2+3^3\right)+.........+\left(3^{10}+3^{11}\right)\)
\(\Leftrightarrow A=1\left(1+3\right)+3^2\left(1+3\right)+.........+3^{10}\left(1+3\right)\)
\(\Leftrightarrow A=1.4+3^2.4+.......+3^{10}.4\)
\(\Leftrightarrow A=4\left(1+3^2+..........+3^{10}\right)⋮4\left(đpcm\right)\)
Bài 1:Ta có:315+314=314.3+314=314.4 chia hết cho 4
Bài 2:a,\(3A=3+3^2+3^3+...........+3^{2016}\)
\(\Rightarrow3A-A=\left(3+3^2+.......+3^{2016}\right)-\left(1+3+.......+3^{2015}\right)\)
\(\Rightarrow2A=3^{2016}-1\Rightarrow A=\frac{3^{2016}-1}{2}\)
b,Ta có:A=1+3+32+33+.............+32015
=(1+3)+(32+33)+...............+(32014+32015)
=4+32.4+................+32014.4
=4.(1+32+.........+32014) chia hết cho 4
A = 3 + 32 + 33 + 34 + .... + 399 + 3100
= (3 + 32 + 33 + 34) + (35 + 36 + 37 + 38) + ..... + (397 + 398 + 399 + 3100)
= 3(1 + 3 + 32 + 33) + 35(1 + 3 + 32 + 33) + .... + 397(1 + 3 + 32 + 33)
= 40(3 + 35 + .... + 397) \(⋮5\)
Ta thấy A \(⋮3\)(vì các số hạng của A đều chia hết cho 3)
mà (3; 5) = 1
nên A \(⋮15\)
Ta có : A =3+3^2+3^3+3^4+.............+3^99+3^100
= (3+3^2+3^3+3^4)+................+(3^97+3^98+3^99+3^100)
= 3.(1+2+3+3^2)+ ...............+3^97.(1+2+3+3^2)
=3.15+.........................+3^97.15
=15.(3+...............+3^97) chia hết cho 15
Muốn chứng minh A thì chúng ta phải tìm A trước :
A = 2.A - A
Tính 2.A = 2 . ( 1 + 32 + 33 + 34 +...+311)
2.A = 2 . ( 1 + 33 + 34 + 35+ ... + 311 + 312 )
Tìm A : A= 2A -A
= ( 1 + 33 + 34 + 35+ ... + 311 + 312 ) - ( 1 + 32 + 33 + 34 +...+311)
= 32 + 312
= 314 = 4782969
4782969 chia hết cho 13 nhưng chia không hết cho 40
\(A=\left(3+3^2+3^3+3^4\right)+3^4\left(3+3^2+3^3+3^4\right)+...+3^{2008}\left(3+3^2+3^3+3^4\right)\)
\(=120+3^4.120+...+3^{2008}.120=120\left(1+3^4+...+3^{2008}\right)⋮120\)
\(A=\left(3+3^2+3^3+3^4\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\)
\(A=\left(3+3^2+3^3+3^4\right)+...+3^{2008}\left(3+3^2+3^3+3^4\right)\)
\(A=\left(3+3^2+3^3+3^4\right)\left(1+3^4+...+3^{2008}\right)\)
\(A=120\left(1+3^4+...+3^{2008}\right)⋮120\)
Ta có: \(A=3+3^2+3^3+...+3^{2012}\)
\(=\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\)
\(=3\left(1+3+3^2+3^3\right)+3^5\left(1+3+3^2+3^3\right)+...+3^{2009}\left(1+3+3^2+3^3\right)\)
\(=3.40+3^5.40+...+3^{2009}.40\)
\(=120+3^4.120+...+3^{2008}.120\)
\(=120\left(1+3^4+...+3^{2008}\right)\)
Vì \(120⋮120\) nên \(120\left(1+3^4+...+3^{2008}\right)⋮120\)
hay \(A⋮120\) (đpcm)
c)D=4+42+43+44+...+42012
D=(4+42)+(43+44)+...+(42011+42012)
D=4.5+43.5+45.5+...+42011.5
D=5.(4+43+42011)
=>D chia hết cho 5
=>ĐPCM
Answer:
\(A=3^1+3^2+3^3+3^4+...+3^{2012}\)
\(=\left(3^1+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2011}+3^{2012}\right)\)
\(=3.\left(1+3\right)+3^3.\left(1+3\right)+...+3^{2011}.\left(1+3\right)\)
\(=4.\left(3+3^2+...+3^{2011}\right)\)
\(=2.2.\left(3+3^3+...+3^{2011}\right)⋮2\)
Vậy \(A⋮2\)