Past Papers’ Solutions  Edexcel  AS & A level  Mathematics  Core Mathematics 1 (C16663/01)  Year 2008  January  Q#7
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Question
A sequence is given by:
x_{1}=1,
x_{n+1}=x_{n}(p+
x_{n}),
where p is a constant (p≠0)
.
a. Find x_{2} in terms of p.
b. Show that x_{3}=1+3p+2p^{2}.
Given that x_{3}=1,
c. find the value of p,
d. (d) write down the value of x_{2008} .
Solution
a.
We are given the sequence as;
x_{1}=1
x_{n+1}=x_{n}(p+ x_{n})
We are required to find x_{2}.
x_{n+1}=x_{n}(p+x_{n})
x_{1+1}=x_{1}(p+x_{1})
x_{2}=(1)(p+1)
x_{2}= p+1
b.
We are required to find x_{3}.
x_{n+1}=x_{n}(p+ x_{n})
x_{2+1}=x_{2}(p+ x_{2})
x_{3}=(p+1)(p+p+1)
x_{3}= (p+1)(2p+1)
x_{3}=(p)(2p)+(p)(1)+ (1)(2p)+ (1)(1)
x_{3}=2p^{2} +p+
2p+1
x_{3}=2p^{2} +3p+1
c.
We have found in (b) that;
x_{3}=2p^{2} +3p+1
We are given that x_{3}=1.
We are required to find p.
Substitute x_{3}=1 in above equation.
1=2p^{2} +3p+1
11=2p^{2} +3p+11
2p^{2} +3p=0
p(2p+3)=0
Now we have two options.
p=0 
2p+3=0 
2p+33=03 

2p=3 




We are given that p≠0, therefore, .
d.
We are given that sequence is;
x_{1}=1
x_{n+1}=x_{n}(p+x_{n})
We have also found in (c) that;
Therefore;
It is evident that all odd terms are the same and similar is the case of even terms.
We are required to find x_{2008} which is an even term, therefore;
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