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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: A = 2 + 22 + 23 + ... + 22008
=> 2A = 22 + 23 + ... + 22009
=> 2A - A = 22009 - 2
=> A = 22009 - 2
Vậy A = 22009 - 2
=> 2A = 2.(22009 - 2)
=> 2A = 22010 - 4
=> 2A + 4 = 22010 - 4 + 4
=> 2A + 4 = 22010
=> 3x = 2010
Sai đề
\(A=2+2^2+2^3+...+2^{2008}\)
\(2A=2^2+2^3+2^4+...+2^{2009}\)
\(2A-A=\left(2^2+2^3+2^4+...+2^{2009}\right)-\left(2+2^2+2^3+...+2^{2008}\right)\)
\(A=2^{2009}-2\)
\(\Rightarrow2.\left(2^{2009}-2\right)+4=2^{3x}\)
\(\Rightarrow2^{2010}-4+4=2^{3x}\)
\(\Rightarrow2^{2010}=2^{3x}\)
=> 3x = 2010
=> x =670
![](https://rs.olm.vn/images/avt/0.png?1311)
A= 3+3^2+3^3+...3^2008
=> 3A = 3^2+3^3+...3^2008+3^2009
=> 3A-A=2A= ( 3^2+3^3+...3^2008+3^2009 ) - ( 3+3^2+3^3+...3^2008 )
=> 2A= 3^2009-3
Mà : 2A+3=3n
=> 3n = 3^2009-3
còn đến đây thì bạn tìm n như tìm x thôi :)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=3+3^2+...+3^{2008}\)
\(\Rightarrow3A=3^2+3^3+...+3^{2009}\)
\(\Rightarrow3A-A=3^{2009}-3\)
\(\Rightarrow2A+3=3^{2009}\)
Vậy n = 2009
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có : \(3A=3^{2007}+3^{2006}+...+3^3+3^2\)
A = \(3^{2006}+...+3^3+3^2+3\)
\(\Rightarrow2A=3^{2007}-3\)
\(\Rightarrow A=\frac{3^{2007}-3}{2}\)
b) Ta có \(2A=3^{2007}-3\)\(\Rightarrow2A+3=3^{2007}\)
Theo bài ta có: \(2A+3=3x\)
\(\Rightarrow3^{2007}=3x\)
\(\Rightarrow3.3^{2006}=3x\)
\(\Rightarrow x=3^{2006}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Mình tưởng là 2A+3=3x chứ bạn?
A=3+32+33+...+3200
3A=32+33+34+...+3201
3A-A=(32+33+34+...+3201)-(3+32+33+...+3200)
2A=3201-3
=>2A+3=3201 =>x=201
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
\(3A=3^2+3^3+....+3^{201}\)
\(\Rightarrow3A-A=2A=\left(3^2+3^3+....+3^{201}\right)-\left(3+3^2+....+3^{200}\right)\)
\(\Rightarrow2A=3^{201}-3\)
\(\Rightarrow2A-3=3^{201}\)
Mà 2A - 3= 3x
=>x=3200
3a=32+33+...+3201
3a-a=2a= (3^2+3^3+...+3^201)-(3+3^2+...+3^201)
2a =3^201 -3
2a-3 =3^201
=>x= 3^200
![](https://rs.olm.vn/images/avt/0.png?1311)
A=3+32+33+...+32009
=>3A=32+33+...+32010
=>3A-A=32+33+...+32010-(3+32+33+...+32009)
=>2A=32+33+...+32010-3-32-33-...-32009
=>2A=33010-3
=>2A+3=32010
mà 2A+3=3n
=>n=2010
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=3+3^2+3^3+...+3^{100}\)
\(\Rightarrow3A=3^2+3^3+3^4+...+3^{101}\)
\(\Rightarrow2A=3A-A=3^2+3^3+...+3^{101}-3-3^2-...-3^{100}=3^{101}-3\)Ta có: \(2A+x=3^{2020}\)
\(\Rightarrow3^{101}-3+x=3^{2020}\)
\(\Rightarrow x=3^{2020}+3-3^{101}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
suy ra 3.A=3^2+...+3^101
3A-A=(3^2+...+3^101)-(3+...+3^100)
2A=3^101-3
A=(3^101-3):2
2A+3=(3^101-3):2.2+3
=3^101-3+3
=3^101
3^x=3^101
Vậy x =101
A= 3+3^2+3^3+...+3^2008
3A=3^2+3^3+3^4+...+3^2008+3^2009
3A - A= (3^2+3^3+3^4+...+3^2008+3^2009)-(3+3^2+3^3+...+3^2008)
2A= 3^2009-3
=>2A+3=3^2009
=>3^x=3^2009
=>x=2009
vậu x= 2009
3.A=3^2+3^3+3^4+...+3^2009
3.A-A=(3^2+3^3+3^4+...+3^2009)-(3+3^2+3^3+...+3^2008)
2.A=3^2009-3
2.A+3=3^2009-3+3
2.A+3=3^2009
đúng k cho mình nhé