Scottish Highers Maths Recurrence Relations: Sequences may be generated by re

Sequences may be generated by recurrence relations of the form $u_{n+1} = k u_{n} - 20$, $u_{0} = 5$ where $k \in \mathbb{R}$ - Scottish Highers Maths - Question 8 - 2017

Question 8

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Sequences may be generated by recurrence relations of the form $u_{n+1} = k u_{n} - 20$, $u_{0} = 5$ where $k \in \mathbb{R}$. (a) Show that $u_{n} = 5k^{n} - 2... show full transcript

Worked Solution & Example Answer:Sequences may be generated by recurrence relations of the form $u_{n+1} = k u_{n} - 20$, $u_{0} = 5$ where $k \in \mathbb{R}$ - Scottish Highers Maths - Question 8 - 2017

Step 1

Show that $u_{n} = 5k^{n} - 20k - 20$

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Answer

To demonstrate that $u_{n} = 5k^{n} - 20k - 20$ , we can use the recurrence relation.
Starting with the initial condition:

Base Case:
For $n=0$ ,
$u_{0} = 5$
This matches with our equation because:
$u_{0} = 5k^{0} - 20k - 20 = 5 - 20k - 20$ which simplifies to $5 = 5$ , confirming the base case.
Inductive Step:
Assume the relation holds for some integer $n$ , that is, assume
$u_{n} = 5k^{n} - 20k - 20$
We need to show it holds for $n+1$ , so substituting into the recurrence relation:

$u_{n+1} = k u_{n} - 20$
Substitute for $u_{n}$ :
$u_{n+1} = k(5k^{n} - 20k - 20) - 20$
Expanding this, we get:
$u_{n+1} = 5k^{n+1} - 20k^{2} - 20k - 20$

The constructed equation matches our original formula as $u_{n+1} = 5k^{n+1} - 20k - 20$ . Thus, the expression is validated for all $n \geq 0$ .

Step 2

Determine the range of values of $k$ for which $u_{n} < u_{0}$

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Answer

We need to analyze the inequality $u_{n} < u_{0}$ where $u_{0} = 5$ .
Substituting our expression for $u_{n}$ gives us:

$5k^{n} - 20k - 20 < 5$
This simplifies to:
$5k^{n} - 20k - 25 < 0$
Dividing through by 5 results in:
$k^{n} - 4k - 5 < 0$
This is a quadratic expression in terms of $k$ . To find the roots, we can set:
$k^{n} - 4k - 5 = 0$
Using the quadratic formula, $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a=1, b=-4, c=-5$ :
$k = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-5)}}{2(1)}$
$k = \frac{4 \pm \sqrt{16 + 20}}{2}$
$k = \frac{4 \pm \sqrt{36}}{2}$
$k = \frac{4 \pm 6}{2}$
Thus, we find two roots:

$k = 5$
$k = -1$
To determine the range, we analyze the intervals defined by these roots.
Testing values across the intervals:

For $k < -1$ , the expression is positive.
For $k = -1$ , the expression equals zero.
For $-1 < k < 5$ , the expression is negative (valid).
For $k = 5$ , the expression equals zero.
For $k > 5$ , the expression is positive.
This leads us to the conclusion that:

The range of values of $k$ for which $u_{n} < u_{0}$ is $-1 < k < 5$ .

Sequences may be generated by recurrence relations of the form $u_{n+1} = k u_{n} - 20$, $u_{0} = 5$ where $k \in \mathbb{R}$ - Scottish Highers Maths - Question 8 - 2017

Worked Solution & Example Answer:Sequences may be generated by recurrence relations of the form $u_{n+1} = k u_{n} - 20$, $u_{0} = 5$ where $k \in \mathbb{R}$ - Scottish Highers Maths - Question 8 - 2017

Show that $u_{n} = 5k^{n} - 20k - 20$

Determine the range of values of $k$ for which $u_{n} < u_{0}$

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