Solution: 2013 Fall Midterm - 11
Author: Pat MorinQuestion
The function $ f: \mathbb{N} \rightarrow \mathbb{N} $ is defined by
$ f(0) = 14 $
$ f(n+1) = f(n) + 4n - 5 $ for $ n \geq 0 $
What is $ f(n) $?
(a)
$f(n) = 2n^{2} - 6n + 14$
(b)
$f(n) = 2n^{2} - 7n + 14$
(c)
$f(n) = 2n^{2} + 7n + 14$
(d)
$f(n) = 2n^{2} + 6n + 14$
Solution
Let’s calculate $ f(1) $
$ f(1) = 14 + 4(0) - 5 = 9 $
- $ f(n) = 2n^{2} + 6n + 14 $
$ f(1) = 2(1)^2 + 6(1) + 14 = 22 $ - $ f(n) = 2n^{2} - 6n + 14 $
$ f(1) = 2(1)^2 - 6(1) + 14 = 10 $ - $ f(n) = 2n^{2} + 7n + 14 $
$ f(1) = 2(1)^2 + 7(1) + 14 = 23 $ - $ f(n) = 2n^{2} - 7n + 14 $
$ f(1) = 2(1)^2 - 7(1) + 14 = 9 $
Since only d) gives us the same output as the original function, we chilling with it fr fr