Solution: 2015 Winter Midterm - 10

Author: Michiel Smid

Question

The function $f: \mathbb{N} \rightarrow \mathbb{N}$ is defined by $$ \begin{align} f(0) &= 15 \\ f(n) &= f(n - 1) +6n - 4\; \ \mathrm{for}\ n \geq 1 \end{align} $$ What is $f(n)$?

(a)

$f(n) = 3n^{2} - 2n + 15$

(b)

$f(n) = 3n^{2} - n + 15$

(c)

$f(n) = 3n^{2} + n + 15$

(d)

$f(n) = 3n^{2} + 2n + 15$

Solution

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

$f(1) = f(0) + 6(1) - 4 = 15 + 6 - 4 = 17$

$f(2) = f(1) + 6(2) - 4 = 17 + 12 - 4 = 25$

$f(n) = 3n^{2} + 2n + 15$
$f(1) = 3{(1)}^{2} + 2(1) + 15 = 3 + 2 + 15 = 20$
$f(n) = 3n^{2} - 2n + 15$
$f(1) = 3{(1)}^{2} - 2(1) + 15 = 3 - 2 + 15 = 16$
$f(n) = 3n^{2} + n + 15$
$f(1) = 3{(1)}^{2} + 1 + 15 = 3 + 1 + 15 = 19$
$f(n) = 3n^{2} - n + 15$
$f(1) = 3{(1)}^{2} - 1 + 15 = 3 - 1 + 15 = 17$

Therefore, we know that $f(n) = 3n^{2} - n + 15$ is the correct answer