cho A = 3^1 + 3^2 +3^3 +3^4+...+3^2012.chứng minh rằng A chia hết cho 120
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
A=1+4+42+...+459A=1+4+42+...+459
A=(1+4+42)+(43+44+45)+...+(457+458+459)A=(1+4+42)+(43+44+45)+...+(457+458+459)
A=(1+4+42)+43(1+4+42)+...+447(1+4+42)A=(1+4+42)+43(1+4+42)+...+447(1+4+42)
A=21+43.21+...+447.21A=21+43.21+...+447.21
A=21(1+43+...+447)A=21(1+43+...+447)
⇒A⋮21
các số như 43,447,459,458........ là 4 mũ và các số đằng sau là số mũ
câu b cũng làm như vậy nhưng dổi các số và kết quả
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn ơi đề thừa số 1 thì phải nha
A = (3+3^2)+(3^3+3^4)+....+(3^2011+3^2012)
= 3.(1+3)+3^3.(1+3)+....+3^2011.(1+3)
= 4+3^3.4+.....+3^2011.4
= 4.(3+3^3+....+3^2011) chia hết cho 4
k mk nha
![](https://rs.olm.vn/images/avt/0.png?1311)
A=(3^0+3^1+3^2+3^3)+(3^4+3^5+3^6+3^7)+...+(3^2009+3^2010+3^2011+3^2012)
A=40+3^4*(1+3+3^2+3^3)+...+3^2009*(1+3+3^2+3^3)
A-1=40+80*40+...+3^2009*40
A-1=40*(1+80+..+3^2009)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có ; \(A=3+3^2+3^3+.....+3^{100}\)
\(=\left(3+3^2+3^3+3^4+3^5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A= 1 +(3^1+3^2+3^3+3^4)+..............................+(3^2009+3^2010+3^2011+3^2012)
A=1+120+................................+3^2009*(3^1+3^2+3^3+3^4)
A=1+(1+.....................+3^2009)*120
Vì 120 chia hết cho 40
suy ra (1+..........................+3^2009) chia hết cho 40
suy ra A chia 40 dư 1
suy ra A-1 chia hết cho 40
![](https://rs.olm.vn/images/avt/0.png?1311)
4a=4+42+43+......+42013
4a-a=(4+42+43+......+42013)-(1+4+42+......+42012)
3a=42013-1
a=42013-1
3
\(A=\left(3+3^2+3^3+3^4\right)+3^4\left(3+3^2+3^3+3^4\right)+...+3^{2008}\left(3+3^2+3^3+3^4\right)\)
\(=120+3^4.120+...+3^{2008}.120=120\left(1+3^4+...+3^{2008}\right)⋮120\)
\(A=\left(3+3^2+3^3+3^4\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\)
\(A=\left(3+3^2+3^3+3^4\right)+...+3^{2008}\left(3+3^2+3^3+3^4\right)\)
\(A=\left(3+3^2+3^3+3^4\right)\left(1+3^4+...+3^{2008}\right)\)
\(A=120\left(1+3^4+...+3^{2008}\right)⋮120\)