Respuesta :
Given:
The rule for the sequence is
[tex]f(n)=3n^2+1[/tex]
Domain of the function is the set of whole numbers greater than 0.
To find:
The first 4 terms of the sequence defined by the given rule.
Solution:
We have,
[tex]f(n)=3n^2+1[/tex]
Domain of the function is the set of whole numbers greater than 0. So, domain for first four terms are 1, 2, 3 and 4 respectively.
For n=1,
[tex]f(1)=3(1)^2+1[/tex]
[tex]f(1)=3(1)+1[/tex]
[tex]f(1)=3+1[/tex]
[tex]f(1)=4[/tex]
For n=2,
[tex]f(2)=3(2)^2+1[/tex]
[tex]f(2)=3(4)+1[/tex]
[tex]f(2)=12+1[/tex]
[tex]f(2)=13[/tex]
For n=3,
[tex]f(3)=3(3)^2+1[/tex]
[tex]f(3)=3(9)+1[/tex]
[tex]f(3)=27+1[/tex]
[tex]f(3)=28[/tex]
For n=4,
[tex]f(4)=3(4)^2+1[/tex]
[tex]f(4)=3(16)+1[/tex]
[tex]f(4)=48+1[/tex]
[tex]f(4)=49[/tex]
Therefore, the first 4 terms of the sequence defined by the given rule are 4, 13, 28 and 49 respectively.
