A sequence of numbers $a_1, a_2, a_3, \, \ldots$ is defined by $$a_{n+1} = \frac{k(a_n + 2)}{a_n}, \quad n \in \mathbb{N}$$ where $k$ is a constant. Given that - the sequence is a periodic sequence of order 3 - $a_1 = 2$ (a) show that $$k^2 + k - 2 = 0$$ (b) For this sequence explain why $k \neq 1$ (c) Find the value of $$\sum_{n=1}^{80} a_n$$

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A sequence of numbers $a_1, a_2, a_3, \, \ldots$ is defined by $$a_{n+1} = \frac{k(a_n + 2)}{a_n}, \quad n \in \mathbb{N}$$ where $k$ is a constant - Edexcel - A-Level Maths Pure - Question 15 - 2020 - Paper 1

Question 15

$A-sequence-of-numbers-$a_1,-a_2,-a_3,-\,-\ldots$-is-defined-by--$$a_{n+1}-=-\frac{k(a_n-+-2)}{a_n},-\quad-n-\in-\mathbb{N}$$--where-$k$-is-a-constant-Edexcel-A-Level Maths Pure-Question 15-2020-Paper 1.png$

A sequence of numbers $a_1, a_2, a_3, \, \ldots$ is defined by $$a_{n+1} = \frac{k(a_n + 2)}{a_n}, \quad n \in \mathbb{N}$$ where $k$ is a constant. Given that - ... show full transcript

Worked Solution & Example Answer:A sequence of numbers $a_1, a_2, a_3, \, \ldots$ is defined by $$a_{n+1} = \frac{k(a_n + 2)}{a_n}, \quad n \in \mathbb{N}$$ where $k$ is a constant - Edexcel - A-Level Maths Pure - Question 15 - 2020 - Paper 1

Step 1

show that $k^2 + k - 2 = 0$

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Answer

To show that $k^2 + k - 2 = 0$ , we start by calculating the first few terms of the sequence using the formula provided.

Finding the first few terms:

Given that $a_1 = 2$ , we can calculate:
- For $n = 1$ : $a_2 = \frac{k(a_1 + 2)}{a_1} = \frac{k(2 + 2)}{2} = \frac{4k}{2} = 2k$
- For $n = 2$ : $a_3 = \frac{k(a_2 + 2)}{a_2} = \frac{k(2k + 2)}{2k} = \frac{2k(k + 1)}{2k} = k + 1$
- For $n = 3$ : $a_4 = \frac{k(a_3 + 2)}{a_3} = \frac{k(k + 1 + 2)}{k + 1} = \frac{k(k + 3)}{k + 1}$
Setting up periodicity condition:

Since the sequence is periodic of order 3, we require that $a_4 = a_1 = 2$ .

Thus, we have: $\frac{k(k + 3)}{k + 1} = 2$

Rearranging gives us: $k(k + 3) = 2(k + 1)$

Expanding both sides: $k^2 + 3k = 2k + 2$

Simplifying further: $k^2 + 3k - 2k - 2 = 0$ $k^2 + k - 2 = 0$

Hence, proved.

Step 2

For this sequence explain why $k \neq 1$

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Answer

If $k = 1$ , we would have:

$a_2 = 2k = 2$
$a_3 = k + 1 = 2$
$a_4 = \frac{k(k + 3)}{k + 1} = \frac{1(1 + 3)}{1 + 1} = 2$

Thus, all terms would equal 2, leading to a constant sequence. A periodic sequence of order 3 must have at least three different values in one complete cycle, indicating that $k$ cannot be equal to 1.

Step 3

Find the value of $\sum_{n=1}^{80} a_n$

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Answer

To find the sum $\sum_{n=1}^{80} a_n$ , we first determine the repeating terms as:

$a_1 = 2$ , $a_2 = -4$ , $a_3 = -1$ . The sequence repeats every three terms.

Therefore, the sum for one complete cycle of three terms is: $2 + (-4) + (-1) = -3$

Since there are $rac{80}{3} = 26$ complete cycles with 2 terms remaining:

Total sum for complete cycles: $26 \times (-3) = -78$

Adding the first two terms of the next cycle ( $a_1 = 2$ , $a_2 = -4$ ): $a_1 + a_2 = 2 - 4 = -2$

Thus, the final result is: $\sum_{n=1}^{80} a_n = -78 + (-2) = -80$

A sequence of numbers $a_1, a_2, a_3, \, \ldots$ is defined by $$a_{n+1} = \frac{k(a_n + 2)}{a_n}, \quad n \in \mathbb{N}$$ where $k$ is a constant - Edexcel - A-Level Maths Pure - Question 15 - 2020 - Paper 1

Worked Solution & Example Answer:A sequence of numbers $a_1, a_2, a_3, \, \ldots$ is defined by $$a_{n+1} = \frac{k(a_n + 2)}{a_n}, \quad n \in \mathbb{N}$$ where $k$ is a constant - Edexcel - A-Level Maths Pure - Question 15 - 2020 - Paper 1

show that $k^2 + k - 2 = 0$

For this sequence explain why $k \neq 1$

Find the value of $\sum_{n=1}^{80} a_n$

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