Which recursive formula can be used to generate the sequence shown, where f(1) = 9.6 and n > 1?

9.6, –4.8, 2.4, –1.2, 0.6, ...

Respuesta :

The correct answer is f(n + 1) = –0.5f(n) since n is greater than 1

Answer:

[tex]f(n)=-0.5\times f(n-1)[/tex]

Step-by-step explanation:

Sequence: 9.6, –4.8 , 2.4, –1.2, 0.6, ...

So, f(1) = first term = 9.6

r = common ratio = [tex]\frac{-4.8}{9.6} =\frac{2.4}{-4.8} = -0.5[/tex]

Now , formula of nth term in G.P. = [tex]f(n)=f(1)\times r^{n-1}[/tex]

So, formula for nth term of the given sequence =  [tex]f(n)=9.6\times (-0.5)^{n-1}[/tex]

So,  [tex]f(n-1)=9.6\times (-0.5)^{n-1-1}[/tex]

      [tex]f(n-1)=9.6\times (-0.5)^{n-2}[/tex]

Recursive formula :

[tex]\frac{f(n)}{f(n-1)}= \frac{9.6\times (-0.5)^{n-1}}{9.6\times (-0.5)^{n-2}}[/tex]

[tex]\frac{f(n)}{f(n-1)}=(-0.5)^{n-1-(n-2)}[/tex]

[tex]\frac{f(n)}{f(n-1)}=(-0.5)^{-1+2}[/tex]

[tex]\frac{f(n)}{f(n-1)}=-0.5[/tex]

[tex]f(n)=-0.5\times f(n-1)[/tex]

Hence recursive formula can be used to generate the sequence shown, where f(1) = 9.6 and n > 1 is [tex]f(n)=-0.5\times f(n-1)[/tex]