Solution: 2019 Winter Midterm - 11

Author: Michiel Smid

Question

The function $f : \mathbb{N} \rightarrow \mathbb{N}$ is recursively defined as follows: $$ \begin{align} f(0) &= 6, \\ f(n) &= 4 \cdot f(n-1) + 2^{n} \ \ \mathrm{if}\ n \geq 1. \end{align} $$ Which of the following is true for all integers $n \geq 0$?

(a)

$f(n) = 6 \cdot 4^{n} - 2^{n}$

(b)

None of the above.

(c)

$f(n) = 7 \cdot 4^{n} - 2^{n}$

(d)

$f(n) = 8 \cdot 4^{n} - 2^{n+1}$

Solution

We can find the first few values of $ f(n) $ to find a pattern:

$ f(1) = 4 \cdot f(0) + 2^1 = 4 \cdot 6 + 2 = 26 $

$ f(2) = 4 \cdot f(1) + 2^2 = 4 \cdot 26 + 4 = 108 $

$ f(n) = 6 \cdot 4^{n} - 2^{n} $
$ f(1) = 6 \cdot 4^{1} - 2^{1} = 24 - 2 = 22 $
$ f(n) = 7 \cdot 4^{n} - 2^{n} $
$ f(1) = 7 \cdot 4^{1} - 2^{1} = 28 - 2 = 26 $
$ f(n) = 8 \cdot 4^{n} - 2^{n+1} $
$ f(1) = 8 \cdot 4^{1} - 2^{2} = 32 - 4 = 28 $

Thus, the correct answer is $ f(n) = 7 \cdot 4^{n} - 2^{n} $.