Write each sequence as a function.c. an+1 = 3an, a1 = 1, where nn is a positive integer greater than or equal to 1.

Answer: [tex]f(n)=3^{n-1}[/tex]Step-by-step explanation:Given Recursive formula : [tex]a_{n+1} = 3a_n,[/tex], [tex]a_1=1[/tex]Then,  [tex]a_2=a_{1+1} = 3a_1=3(1)=3[/tex][tex]a_3=a_{2+1} = 3a_2=3(3)=9[/tex][tex]a_4=a_{3+1} = 3a_3=3(9)=27[/tex]We can write it as : [tex]f(n)=3^{n-1}[/tex]such thatn      [tex]f(n)=3^{n-1}[/tex]1       [tex]f(1)=3^{1-1}=1[/tex]2       [tex]f(2)=3^{2-1}=3^1=3[/tex]3       [tex]f(3)=3^{3-1}=3^2=9[/tex]Hence, the required function: [tex]f(n)=3^{n-1}[/tex]