A sequence is given by: $x_1 = 1,$ $x_{n+1} = x_n(p + x_n)$, where $p$ is a constant ($p \neq 0$) - Edexcel - A-Level Maths Pure - Question 9 - 2008 - Paper 2

From the previous calculation in part (a), we derived:

$x_3 = 1 + 3p + 2p^2$

This confirms the result given in the question.

We are given that $x_1 = 1$ and we have just derived:

$x_3 = 1 + 3p + 2p^2$

Next, we need to find $p$. From the previous steps, we know that:

1. Set $x_3 = x_1 = 1$: $1 = 1 + 3p + 2p^2$ Rearranging gives: $0 = 3p + 2p^2$

2. Factor the equation: $0 = p(3 + 2p)$ The solutions are:

   • $p = 0$ (which we discard since $p \neq 0$)
   • $3 + 2p = 0 \Rightarrow 2p = -3 \Rightarrow p = -\frac{3}{2}$

Thus, the value of $p$ is: $p = -\frac{3}{2}$

Given that every term in the sequence is generated via the same recurrence relation, it can be observed that:

Since the sequence is defined recursively and all the terms will be the same as $x_1$, we have:

1. Calculate $x_2$, $x_3$, and onwards: Each term remains the same given the nature of the recurrence relation.

Thus, $x_{2008} = x_1 = 1$