Hi good people!,

How do I go about finding the general term for the following sequence:

\({1\over3};{8\over9};1;{64\over81}\)

I have tested the sequence for Arithmetic, Geometric and Quadratic...but cannot determine which...please help...

juriemagic Jul 25, 2018

#1**+1 **

Try the following:

\(1^3\frac{1}{3}; 2^3\frac{1}{3}^2;3^3\frac{1}{3}^3;4^3\frac{1}{3}^4;...etc\)

.Alan Jul 25, 2018

#2**+1 **

**Hi good people!, How do I go about finding the general term for the following sequence: {1\over3};{8\over9};1;{64\over81}**

\(\displaystyle {1\over3};{8\over9};1;{64\over81}\)

\(\huge{ \begin{array}{|rcll|} \hline a_1 &=& {1\over3} \\\\ a_2 &=& {8\over9} \\\\ a_3 &=& 1 \\\\ a_4 &=& {64\over81} \\\\ \ldots \\\\ a_n &=& \dfrac{n^3}{3^n} \\ \hline \end{array} }\)

heureka Jul 25, 2018

#3**0 **

Hi Heureka,

I have no idea why the answer is what you say it is...How you get to it....thanx anyways...

Guest Jul 25, 2018

#3**0 **

Hi Heureka,

I have no idea why the answer is what you say it is...How you get to it....thanx anyways...

Guest Jul 25, 2018

#7**+1 **

**I have no idea why the answer is what you say it is...How you get to it....thanx anyways...**

\(\huge{ \begin{array}{|lrlrlrlrlrl|} \hline & a_{\color{red}1} &;& a_{\color{red}2}&;& a_{\color{red}3} &;& a_{\color{red}4} &;& \ldots &;& a_{\color{red}n} \\ \hline & \dfrac{1}{3}&;& \dfrac{8}{9}&;& 1&;& \dfrac{64}{81}&;& \ldots &;& \\\\ \Rightarrow & \dfrac{1^1}{3^1}&;& \dfrac{2^3}{3^2}&;& 1&;& \dfrac{4^3}{3^4}&;& \ldots &;& \\\\ \Rightarrow & \dfrac{{\color{red}1}^1}{3^{\color{red}1}}&;& \dfrac{{\color{red}2}^3}{3^{\color{red}2}}&; & \dfrac{{\color{red}3}^3}{3^{\color{red}3}}&;& \dfrac{{\color{red}4}^3}{3^{\color{red}4}}&;& \ldots &;&\dfrac{{\color{red}n}^3}{3^{\color{red}n}} \\ \hline \end{array} }\)

heureka
Jul 25, 2018