Given:
[tex]f(1)=3,f(n)=2\times f(n-1)[/tex] ...(i)
For [tex]n\geq 2[/tex].
To find:
The nth term for the given recursive formula.
Solution:
The given recursive formula is of the form of
[tex]f(n)=r\times f(n-1)[/tex] ...(ii)
It is the recursive formula of a GP, where r is common ratio.
On comparing (i) and (ii), we get
[tex]r=2[/tex]
Now,
First term: [tex]a=f(1)=3[/tex]
Common difference: [tex]r=2[/tex]
nth term of a GP is
[tex]f(n)=ar^{n-1}[/tex]
Putting a=3 and r=2, we get
[tex]f(n)=3(2)^{n-1}[/tex]
Therefore, the equation for the nth term is [tex]f(n)=3(2)^{n-1}[/tex].
