Answer:
See below
Step-by-step explanation:
Let s first find a simplified expression for the sum of the first n terms of the geometric sequence [tex]\bf \left\{\displaystyle\frac{1}{3^k}\right\}_{k=1}^n[/tex]
[tex]\bf S_n=\displaystyle\frac{1}{3}+\displaystyle\frac{1}{3^2}+\displaystyle\frac{1}{3^3}+...+\displaystyle\frac{1}{3^n}[/tex]
If we multiply both sides by 1/3, we get
[tex]\bf \displaystyle\frac{1}{3}S_n=\displaystyle\frac{1}{3^2}+\displaystyle\frac{1}{3^3}+...+\displaystyle\frac{1}{3^n}+\displaystyle\frac{1}{3^{n+1}}[/tex]
Hence
[tex]\bf S_n-\displaystyle\frac{1}{3}S_n=\displaystyle\frac{1}{3}-\displaystyle\frac{1}{3^{n+1}}\Rightarrow \displaystyle\frac{2}{3}S_n=\displaystyle\frac{1}{3}-\displaystyle\frac{1}{3^{n+1}}\Rightarrow\\\\\Rightarrow 2S_n=1-\displaystyle\frac{1}{3^n}=\displaystyle\frac{3^n-1}{3^n}\Rightarrow S_n=\displaystyle\frac{3^n-1}{2(3^n)}[/tex]
Now, we have
[tex]\bf a_1=6\\\\a_2=6+\displaystyle\frac{1}{3}=\displaystyle\frac{19}{3}\\\\a_3=6+\displaystyle\frac{1}{3}+\displaystyle\frac{1}{3^2}=6+S_2=6+\displaystyle\frac{8}{18}=\displaystyle\frac{58}{9}\\\\a_4=6+S_3=6+\displaystyle\frac{26}{54}=\displaystyle\frac{175}{27}[/tex]
[tex]\bf a_5=6+S_4=6+\displaystyle\frac{3^4-1}{2(3^4)}=\displaystyle\frac{526}{81}\\\\a_6=6+S_5=6+\displaystyle\frac{3^5-1}{2(3^5)}=\displaystyle\frac{1579}{243}\\\\a_7=6+S_6=6+\displaystyle\frac{3^6-1}{2(3^6)}=\displaystyle\frac{4738}{729}[/tex]
[tex]\bf a_8=6+S_7=6+\displaystyle\frac{3^7-1}{2(3^7)}=\displaystyle\frac{14215}{2187}\\\\a_9=6+S_8=6+\displaystyle\frac{3^8-1}{2(3^8)}=\displaystyle\frac{42646}{6561}\\\\a_{10}=6+S_9=6+\displaystyle\frac{3^9-1}{2(3^9)}=\displaystyle\frac{127939}{19683}[/tex]
