Answer :
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 that
n [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]
Answer:
[tex]f(x)=3^x[/tex] where [tex]0\leq x[/tex]
Step-by-step explanation:
We are given that a sequence
[tex]a_{n+1}=3a_n,a_1=1[/tex]
Where [tex]n\geq 1[/tex]
We have to write a sequence as function.
When n=1
[tex]a_2=3a_1[/tex]
[tex]a_2=3(1)=3[/tex]
n=2
[tex]a_3=3a_2=3(3)=9=3^2[/tex]
[tex]a_4=3a_3=3(9)=27=3^3[/tex]
[tex]a_5=3(27)=81=3^4[/tex]
Therefore, the given sequence can be write as function
[tex]f(x)=3^x[/tex] where [tex]0\leq x[/tex]