a,Ta có:3A=3²+3³+................+3²⁰¹¹

\(\Rightarrow3A-A=\left(3^2+3^3+.....+3^{2011}\right)-\left(3+3^2+.....+3^{2010}\right)\)

\(\Rightarrow2A=3^{2011}-3\)

\(\Rightarrow A=\frac{3^{2011}-3}{2}\)

b,Ta có:\(2A=3^{2011}-3\Rightarrow2A+3=3^{2011}\Rightarrow x=2011\)

a) \(A=3^1+3^2+3^3+...+3^{2010}\)

\(3A=3.\left(3^1+3^2+3^3+...+3^{2010}\right)\)

\(3A=3.3^1+3.3^2+3.3^3+...+3.3^{2010}\)

\(3A=3^2+3^3+3^4+...+3^{2011}\)

\(3A-A=2A\)

\(2A=\left(3^2+3^3+3^4+...+3^{2011}\right)-\left(3^1+3^2+3^3+...+3^{2010}\right)\)

\(2A=3^{2011}-3^1=3^{2011}-3\)\(\Rightarrow\)\(A=\left(3^{2011}-3\right)\div2\)

b) Mình ko biết

\(A=3^1+3^2+3^3+...+3^{2010}\)

\(\Rightarrow3A=3^2+3^3+...+3^{2011}\)

\(\Rightarrow2A=3^{2011}-3\)

\(\Rightarrow A=\frac{3^{2011}-2}{2}\)

\(\Leftrightarrow2A+3=3^{2011}-3+3=2^{2011}\)

\(\Rightarrow x=2011\)

a)

Ta có 3A=\(3^2+3^3+3^4+...+3^{2017}\)

3A-A=\(\left(3^2+3^3+3^4+...+3^{2017}\right)-\left(3+3^2+3^3+...+3^{2016}\right)\)

2A=\(3^{2017}-3\)

A=\(\frac{3^{2017}-3}{2}\)

b)

A=\(\frac{3^{2017}-3}{2}\)

2A=\(3^{2017}-3\)

2A+3=\(3^{2017}-3+3=3^{2017}\)

=>x=2017

Bài 1:

a,\(A=3+3^2+3^3+...+3^{2010}\)

\(=\left(3+3^2+3^3+3^4\right)+....+\left(3^{2007}+3^{2008}+3^{2009}+3^{2010}\right)\)

\(=3\left(1+3+3^2+3^3\right)+....+3^{2007}\left(1+3+3^2+3^3\right)\)

\(=3.40+...+3^{2007}.40\)

\(=40\left(3+3^5+...+3^{2007}\right)⋮40\)

Vì A chia hết cho 40 nên chữ số tận cùng của A là 0

b,\(A=3+3^2+3^3+...+3^{2010}\)

\(3A=3^2+3^3+...+3^{2011}\)

\(3A-A=\left(3^2+3^3+...+3^{2011}\right)-\left(3+3^2+3^3+...+3^{2010}\right)\)

\(2A=3^{2011}-3\)

\(2A+3=3^{2011}\)

Vậy 2A+3 là 1 lũy thừa của 3

3A=3^2+3^3+...+3^2007

=>3a-A=(3^2+3^3+...+3^2007)-(3^1+3^2+...+3^2006)

=>2A=3^2007-3^1=3^2007-3

=>2A+3=3^2007-3+3=3^2007=3^x

=>x=2007