$a)$  

$a_n=2^n$  

$a_1=2$  

$a_{n+1}=2^{n+1}$  

 

$\frac{a_{n+1}}{a_n}=\frac{2^{n+1}}{2^n}=2$    premium} 

$\{\begin{array}{l}{a}_1=2\\ a_{n+1}=2a_n\text{,}n\ge 1\end{array}$  

 

$b)$  

$a_n=\frac{1}{n}$  

$a_1=1$  

$a_{n+1}=\frac{1}{n+1}$  

 

$\frac{a_{n+1}}{a_n}=\frac{1}{n+1}\cdot n=\frac{n}{n+1}$  

$\{\begin{array}{l}{a}_1=1\\ a_{n+1}=\frac{n}{n+1}\cdot a_n\text{,}n\ge 1\end{array}$  

 

$c)$  

$a_n=n^2$  

$a_1=1$  

$a_{n+1}={\left(n+1\right)}^2$  

$a_{n+1}-a_n=n^2+2n+1-n^2=2n+1$  

$\{\begin{array}{l}{a}_1=1\\ a_{n+1}=a_n+2n+1\text{,}n\ge 1\end{array}$  

 

$d)$  

$a_n=1+4+9+\dots +n^2$  

$a_1=1$  

$a_{n+1}=1+4+9+\dots +n^2+{\left(n+1\right)}^2$    

$a_{n+1}-a_n={\left(n+1\right)}^2$  

 

$\{\begin{array}{l}{a}_1=1\\ a_{n+1}=a_n+{\left(n+1\right)}^2\text{,}n\ge 1\end{array}$