\(A=3+3^2+3^3+...+3^{120}\)

\(\Rightarrow3A=3\left(3+3^2+3^3+...+3^{100}\right)\)

\(3A=3^2+3^3+3^4+...+3^{101}\)

\(\Rightarrow3A-A=\left(3^2+3^3+...+3^{101}\right)-\left(3+3^3+...+3^{100}\right)\)

\(\Rightarrow2A=3^{101}-3\)

\(\Rightarrow2A+3=3^{101}-3+3=3^{101}=3^n\)

\(\Rightarrow n=101\)

vậy ...

\(3A=3^2+3^3+...+3^{121}\)

\(3A-A=\left(3^2-3^2\right)+........+\left(3^{120}-3^{120}\right)+3^{121}-3\)

A = \(\frac{3^{121}-3}{2}\)

2A + 3 = \(\frac{3^{121}-3}{2}.2+3=3^{121}=3^n\)

Vậy n = 121       

n=121

Ta có: \(A=3+3^2+...+3^{101}\)

\(\Leftrightarrow3A=3^2+3^3+...+3^{102}\)

\(\Rightarrow3A-A=\left(3^2+3^3+...+3^{102}\right)-\left(3+3^2+...+3^{101}\right)\)

\(\Leftrightarrow2A=3^{102}-3\)

\(\Leftrightarrow3^{2n}=2A+3=3^{102}\)

\(\Rightarrow2n=102\)

\(\Rightarrow n=51\)

\(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^3+...+\left(\frac{1}{3}\right)^{2013}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2013}}\)

\(\Rightarrow3B=3\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2013}}\right)\)

\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2012}}\)

\(\Rightarrow3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2012}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2013}}\right)\)

\(\Rightarrow2B=1-\frac{1}{3^{2013}}\Rightarrow1-2B=\frac{1}{3^{2013}}=\left(\frac{1}{3}\right)^{2013}\Rightarrow n=2013\)