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
Question 9
A sequence is given by:
$x_1 = 1,$
$x_{n+1} = x_n(p + x_n),$
where $p$ is a constant $(p \neq 0)$.
(a) Find $x_3$, in terms of $p$.
(b) Show that $x_3 = 1 + 3p... show full transcript
Worked Solution & Example Answer: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
Step 1
Find $x_3$, in terms of $p$
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Answer
To find x3, we start by calculating x2 using the recurrence relation:
For n=1:
x2=x1(p+x1)=1(p+1)=p+1
For n=2:
x3=x2(p+x2)=(p+1)(p+(p+1))=(p+1)(2p+1).
Expanding this gives:
x3=2p2+3p+1.
Step 2
Show that $x_3 = 1 + 3p + 2p^2$
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Answer
From the previous step, we derived:
x3=2p2+3p+1.
This can be rearranged as:
x3=1+3p+2p2.
Hence, the equation is verified.
Step 3
find the value of $p$
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Answer
Given x1=1 and from our expression for x3:
Setting x3=1+3p+2p2=1 leads to:
3p+2p2=0.
Factoring out p, we get:
p(2p+3)=0.
Therefore, p=0 or p=−23. Since p=0, thus:
$$p = -\frac{3}{2}.$
Step 4
write down the value of $x_{2008}$
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Answer
Noting that x2008 follows the same recurrence relation, we see that all even terms are the same:
x2008=x3=1+3(−23)+2(−23)2.
Calculating this gives:
x2008=1−29+2×49=1−29+29=1.
Therefore:
x2008=1.
