Recursion: $f(n+1)=-2f(n)$

Question

Given the function defined recursively as:

$f(n+1)=-2f(n)$

$f(0)=3$

How can I find $f(5)$ using the recursion?

Answer 1

Hint. One has recursively $$ f(5)=-2\times f(4)=-2\times (-2) \times f(3)= \cdots. $$ Hope you can finish it.

Answer 2

You know $f(0) =3$, now us this and apply $f(n+1) = -2f(n)$ to find f(1).

i.e. $f(1) = f(0+1) = -2f(0) = -2(3) = -6$.

repeat this process using $f(2) = f(1+1)$ and so on until you get to

$f(5) = f(4+1) = -2f(4)$

Answer 3

In general, if $f(n+1) = af(n)$, then, for any $k \ge 0$, we have $f(n+k) =a^k f(n)$.

Setting $n=0$, $f(k) = a^kf(0)$.

In your case $a=-2$ and $f(0) = 3$ so $f(k)=3(-2)^k$.

If $k=5$, $f(5) =3(-2)^5 =-3\cdot 32 =-96 $.