Solution: 2016 Fall Midterm - 10

Author: Michiel Smid

Question

The function $f : \mathbb{N} \rightarrow \mathbb{R}$ is defined by $$ \begin{align} f(0) &= 7, \\ f(n) &= \frac{n}{3} \cdot f(n - 1)\; \ \mathrm{for}\ n \geq 1. \end{align} $$ What is $f(n)$?

(a)

$f(n) = 7^n \cdot \frac{(n + 1)!}{3^n}$

(b)

$f(n) = 7^n \cdot \frac{n!}{3^n}$

(c)

$f(n) = 7 \cdot \frac{n!}{3^n}$

(d)

$f(n) = 7 \cdot \frac{(n + 1)!}{3^n}$

Solution

We can calculate values of $ f(1) $ and $ f(2) $ to see if we can find a pattern.

$ f(1) = \frac{1}{3} \cdot 7 = \frac{7}{3} $

$ f(2) = \frac{2}{3} \cdot \frac{7}{3} = \frac{14}{9} $

$ f(n) = 7 \cdot \frac{n!}{3^{n}} $
$ f(1) = 7 \cdot \frac{1!}{3^{1}} = 7 \cdot \frac{1}{3} = \frac{7}{3} $
$ f(2) = 7 \cdot \frac{2!}{3^{2}} = 7 \cdot \frac{2}{9} = \frac{14}{9} $
$ f(n) = 7^{n} \cdot \frac{n!}{3^{n}} $
$ f(1) = 7^{1} \cdot \frac{1!}{3^{1}} = 7 \cdot \frac{1}{3} = \frac{7}{3} $
$ f(2) = 7^{2} \cdot \frac{2!}{3^{2}} = 49 \cdot \frac{2}{9} = \frac{98}{9} $
$ f(n) = 7 \cdot \frac{(n+1)!}{3^{n}} $
$ f(1) = 7 \cdot \frac{2!}{3^{1}} = 7 \cdot \frac{2}{3} = \frac{14}{3} $
$ f(n) = 7^{n} \cdot \frac{(n+1)!}{3^{n}} $
$ f(1) = 7^{1} \cdot \frac{2!}{3^{1}} = 7 \cdot \frac{2}{3} = \frac{14}{3} $

Thus, we know that $ f(n) = 7 \cdot \frac{n!}{3^{n}} $ is the correct answer.