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

\(A=3+3^2+3^3+...+3^{2015}\)

\(\Rightarrow3A=3^2+3^3+...+3^{2015}+3^{2016}\)

\(\Rightarrow3A-A=\left(3^2+3^3+...+3^{2016}\right)-\left(3+3^2+3^3+...+3^{2015}\right)\)

\(\Rightarrow2A=\left(3^2-3^2\right)+\left(3^3-3^3\right)+...+\left(3^{2016}-3\right)\)

\(\Rightarrow2A=3^{2016}-3\)

\(\Rightarrow A=\dfrac{3^{2016}-3}{2}\)

Ta có: \(2A+3=3^n\)

\(\Rightarrow2\cdot\dfrac{3^{2016}-3}{2}+3=3^n\)

\(\Rightarrow3^{2016}-3+3=3^n\)

\(\Rightarrow3^{2016}=3^n\)

\(\Rightarrow n=2016\)

Ở dưới câu của bn

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-Ta có:1+2+3+.........+2006=(2006+1).2006:2=2013021

A=3¹+

a: Tổng các số hạng là:

\(\dfrac{\left(220+1\right)\cdot220}{2}=24310\)

Ta có: A+1=2x

\(\Leftrightarrow2x=24311\)

hay \(x=\dfrac{24311}{2}\)

\(E=\left(1^2+2^2+...+2021^2\right)\left(93-93\right)=0\)

\(a,A=3+3^2+3^3+3^4+...+3^{100}\\ 3A=3^2+3^3+3^4+3^5+3^{101}\\ 3A-A=2A=3^{101}-3\\ \Rightarrow2A+3=3^{101}=3^{4.25+1}\\ \Rightarrow n=25\)

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