Solution: 2016 Fall Final - 8

Author: Michiel Smid

Question

Consider the following recursive function:

f(0) = $1$,
f(n) = $\frac{5}{n} \cdot f(n - 1)\; \ \text{for all}$ $\mathrm{integers}\ n \geq 1$.

Which of the following is true for all $n \geq 0$?

(a)

$f(n) = \frac{5^{n}}{(n+1)!}$

(b)

$f(n) = \frac{5}{n!}$

(c)

$f(n) = \frac{5^{n+1}}{n!}$

(d)

$f(n) = \frac{5^{n}}{n!}$

Let’s calculate $f(1)$ and $f(2)$

$ f(1) = \frac{5}{1} \cdot f(0) = 5 \cdot 1 = 5 $

$ f(2) = \frac{5}{2} \cdot f(1) = \frac{5}{2} \cdot 5 = \frac{25}{2} $

$ f(n) = \frac{5}{n!} $
$ f(1) = \frac{5}{1!} = 5 $
$ f(2) = \frac{5}{2!} = \frac{5}{2} $
$ f(n) = \frac{5^{n}}{n!} $
$ f(1) = \frac{5^{1}}{1!} = 5 $
$ f(2) = \frac{5^{2}}{2!} = \frac{25}{2} $
$ f(n) = \frac{5^{n}}{(n+1)!} $
$ f(1) = \frac{5^{1}}{2!} = \frac{5}{2} $
$ f(2) = \frac{5^{2}}{3!} = \frac{25}{6} $
$ f(n) = \frac{5^{n+1}}{n!} $
$ f(1) = \frac{5^{1+1}}{1!} = 25 $
$ f(2) = \frac{5^{2+1}}{2!} = \frac{125}{2} $

Since b is the only one that has the correct function, it is the answer