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A sequence is defined by the recurrence relation $u_{n+1} = \frac{1}{3}u_n + 10$ with $u_1 = 6$ - Scottish Highers Maths - Question 3 - 2016

Question 3

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A sequence is defined by the recurrence relation $u_{n+1} = \frac{1}{3}u_n + 10$ with $u_1 = 6$. (a) Find the value of $u_6$. (b) Explain why this sequence approac... show full transcript

Worked Solution & Example Answer:A sequence is defined by the recurrence relation $u_{n+1} = \frac{1}{3}u_n + 10$ with $u_1 = 6$ - Scottish Highers Maths - Question 3 - 2016

Step 1

Find the value of $u_6$

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Answer

To find the value of $u_6$ , we will recursively apply the recurrence relation:

Start with $u_1 = 6$ .
Calculate $u_2$ :
$u_2 = \frac{1}{3} u_1 + 10 = \frac{1}{3} \cdot 6 + 10 = 2 + 10 = 12.$
Calculate $u_3$ :
$u_3 = \frac{1}{3} u_2 + 10 = \frac{1}{3} \cdot 12 + 10 = 4 + 10 = 14.$
Calculate $u_4$ :
$u_4 = \frac{1}{3} u_3 + 10 = \frac{1}{3} \cdot 14 + 10 = \frac{14}{3} + 10 = \frac{14}{3} + \frac{30}{3} = \frac{44}{3} \approx 14.67.$
Calculate $u_5$ :
$u_5 = \frac{1}{3} u_4 + 10 = \frac{1}{3} \cdot \frac{44}{3} + 10 = \frac{44}{9} + \frac{90}{9} = \frac{134}{9} \approx 14.89.$
Calculate $u_6$ :
$u_6 = \frac{1}{3} u_5 + 10 = \frac{1}{3} \cdot \frac{134}{9} + 10 = \frac{134}{27} + \frac{270}{27} = \frac{404}{27} \approx 14.96.$
Thus, the value of $u_6$ is approximately $14.96$ .

Step 2

Explain why this sequence approaches a limit as $n \to \infty$

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Answer

The sequence approaches a limit as $n \to \infty$ because the recurrence relation is linear and the coefficient of $u_n$ is less than 1 (specifically, $\frac{1}{3}$ ). This means the influence of the initial condition $u_1$ diminishes as $n$ increases, allowing the sequence to stabilize around a certain value. The addition of the constant term (10) also contributes to this stabilization.

Step 3

Calculate this limit

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Answer

To find the limit, denote it by $L$ . As $n$ approaches infinity, both $u_n$ and $u_{n+1}$ will approach $L$ . Therefore, we can set up the equation:

$L = \frac{1}{3}L + 10.$

To solve for $L$ , we rearrange the equation:

$L - \frac{1}{3}L = 10,$
$\frac{2}{3}L = 10,$
$L = 10 \cdot \frac{3}{2} = 15.$

Thus, the limit of the sequence as $n \to \infty$ is $15$ .

A sequence is defined by the recurrence relation $u_{n+1} = \frac{1}{3}u_n + 10$ with $u_1 = 6$ - Scottish Highers Maths - Question 3 - 2016

Worked Solution & Example Answer:A sequence is defined by the recurrence relation $u_{n+1} = \frac{1}{3}u_n + 10$ with $u_1 = 6$ - Scottish Highers Maths - Question 3 - 2016

Find the value of $u_6$

Explain why this sequence approaches a limit as $n \to \infty$

Calculate this limit

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