Solution: 2017 Winter Midterm - 12

Author: Michiel Smid

Question

The function $f : \{1,2,3,\dots\} \rightarrow \mathbb{R}$ is defined by $$ \begin{align} f(1) &= 2, \\ f(n) &= 2 \left( \frac{n - 1}{n} \right)^2 \cdot f(n - 1)\ \mathrm{if}\ n \geq 2. \end{align} $$ What is $f(n)$?

(a)

$f(n) = \frac{2^{n-1}}{n^{2}}$

(b)

$f(n) = \frac{2^{n}}{n^{2}}$

(c)

$f(n) = \frac{2^{n}}{(n-1)^{2}}$

(d)

$f(n) = \frac{n^{2}}{2^{n}}$

Solution

Let’s find the first few values of $ f(n) $

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

$ f(3) = 2 {( \frac{2}{3})}^{2} \cdot 1 = \frac{8}{9} $

$ \frac{2^{n}}{{(n-1)}^{2}} $
$f(2) = \frac{2^{2}}{{(2-1)}^{2}} = 4 $
$ \frac{2^{n-1}}{n^{2}} $
$f(2) = \frac{2^{1}}{2^{2}} = \frac{1}{2} $
$ \frac{2^{n}}{n^{2}} $
$f(2) = \frac{2^{2}}{2^{2}} = 1 $
$f(3) = \frac{2^{3}}{3^{2}} = \frac{8}{9} $
$ \frac{n^{2}}{2^{n}} $
$f(2) = \frac{2^{2}}{2^{2}} = 1 $
$f(3) = \frac{3^{2}}{2^{3}} = \frac{9}{8} $

Thus, $ \frac{2^{n}}{n^{2}} $ is the correct answer.