Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2008 | January | Q#7

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Question

A sequence is given by:

x1=1,

xn+1=xn(p+
xn),

where p is a constant (p≠0)
.

a.   Find x2 in terms of p.

b.   Show that x3=1+3p+2p2.

Given that x3=1,

c.   find the value of p,

d.    (d) write down the value of x2008 .

Solution

a.
 

We are given the sequence as;

x1=1

xn+1=xn(p+ xn)

We are required to find x2.

xn+1=xn(p+xn)

x1+1=x1(p+x1)

x2=(1)(p+1)

x2= p+1

b.    

We are required to find x3.

xn+1=xn(p+ xn)

x2+1=x2(p+ x2)

x3=(p+1)(p+p+1)

x3= (p+1)(2p+1)

x3=(p)(2p)+(p)(1)+ (1)(2p)+ (1)(1)

x3=2p2 +p+
2p+1

x3=2p2 +3p+1

c.    

We have found in (b) that;

x3=2p2 +3p+1

We are given that x3=1.

We are required to find p.

Substitute x3=1 in above equation.

1=2p2 +3p+1

1-1=2p2 +3p+1-1

2p2 +3p=0

p(2p+3)=0

Now we have two options.

p=0

2p+3=0

2p+3-3=0-3

2p=-3

We are given that p≠0, therefore, .

 

d.
 

We are given that sequence is;

x1=1

xn+1=xn(p+xn)

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 x2008 which is an even term, therefore;

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