<p>A sequence is given by:</p>
<p style="text-align: center">${{x}_{1}}=1$,</p>
<p style="text-align: center">${{x}_{n+1}}={{x}_{n}}(p+{{x}_{n}})$,</p>
<p>where $p$ is a constant ($p\ne 0$).</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Find ${{x}_{2}}$ in terms of $p$.</div><p style="text-align: right;">[1]</p></li><li><div class="list-item-paragraph">Show that ${{x}_{3}}=1+3p+2{{p}^{2}}$.</div><p style="text-align: right;">[2]</p></li></ol>
<p>Given that ${{x}_{3}}=1$,</p>
<ol class="abc-list" start="3"><li><div class="list-item-paragraph">find the value of $p$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Write down the value of ${{x}_{2008}}$.</div><p style="text-align: right;">[2]</p></li></ol>

Recurrence Relations and Summation

International Baccalaureate Higher Level Mathematics

IB HL Math

2008 A-Level Edexcel C1 January Q7

$x_{n+1} = x_n(p + x_n)$

For $n = 1$: $x_2 = x_1(p + x_1)$, given $x_1 = 1$

$x_2 = 1(p + 1) = p + 1$

For $n = 2$: $x_3 = x_2(p + x_2)$

$= (p+1)(p + p+1)$

$= (p+1)(2p+1)$

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

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

Given $x_3 = 1$:

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

$0 = 3p + 2p^2$

$0 = p(3 + 2p)$

$p = 0$ or $p = -\frac{3}{2}$

(reject $p = 0$ as $p \neq 0$)

$\therefore p = -\frac{3}{2}$

Using a calculator with $p = -\frac{3}{2}$:

$x_1 = 1, \quad x_2 = -\frac{1}{2}, \quad x_3 = 1, \quad x_4 = -\frac{1}{2}, \ldots$

The sequence alternates between $1$ and $-\frac{1}{2}$.

Since 2008 is even:

$\therefore x_{2008} = -\frac{1}{2}$

2008 A-Level Edexcel C1 January Q7

2008 A-Level Edexcel C1 January Q7

Video Solution: