Solution: 2015 Fall Midterm - 10

Author: Michiel Smid

Question

The function $f : \mathbb{N} \rightarrow \mathbb{N}$ is defined by $$ \begin{align} f(0) &= 7 \\ f(n) &= f(n - 1) + 10n - 6\; \ \mathrm{for}\ n \geq 1 \end{align} $$ What is $f(n)$?

(a)

$f(n) = 4n^2 - n + 7$

(b)

$f(n) = 5n^2 - 2n + 7$

(c)

$f(n) = 5n^2 - n + 7$

(d)

$f(n) = 4n^2 - 2n + 7$

Solution

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

$f(1) = f(0) + 10(0) - 6 = 7 + 10 - 6 = 11$

$f(2) = f(1) + 10(1) - 6 = 11 + 10 - 6 = 15$

$f(n) = 4n^{2} - 2n + $
$f(1) = 4{(1)}^{2} - 2(1) + 7 = 4 - 2 + 7 = 9$
$f(n) = 4n^{2} - n + 7$
$f(1) = 4{(1)}^{2} - 1 + 7 = 4 - 1 + 7 = 10$
$f(n) = 5n^{2} - 2n + 7$
$f(1) = 5{(1)}^{2} - 2(1) + 7 = 5 - 2 + 7 = 10$
$f(n) = 5n^{2} - n + 7$
$f(1) = 5{(1)}^{2} - 1 + 7 = 5 - 1 + 7 = 11$

Based on the pattern, we can see that $f(n) = 5n^{2} - n + 7$ is the correct answer