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a level maths help exam paper question

1)a sequence is given by
x1=1
xn+1=xn(p+xn)
where p is a constant (not equal to 0)
given that x3=1

d) write down the value of x2008
Reply 1
Avatar for gdunne42
gdunne42
21
sammy3000
1)a sequence is given by
x1=1
xn+1=xn(p+xn)
where p is a constant (not equal to 0)
given that x3=1

d) write down the value of x2008


I assume you have already determined the value of p
now work out the value of the first 4 terms and see what pattern emerges
Reply 2
Avatar for sammy3000
sammy3000
OP
x1=p+1
x2=p^2+1+2p
x3=1+3p+2p^2
Reply 3
Avatar for sammy3000
sammy3000
OP
is there a p being added on.
Reply 4
Avatar for DFranklin
DFranklin
18
Try again. Read the question carefully...
Reply 5
Avatar for paronomase
paronomase
If you have not determined p, I am in a happy mood so...:wink:

By definition, you have x(n+1) = x(n)[p+x(n)]
so using n=2, x(2+1)=x(2)(p+x(2)

Then you change p(2) to p(1+1) and then with your previous formula, you have

= [x(1)(p+x(1)]*(p+[x(1)(p+x(1)]
= (p+1)*(2p+1)
= 2p² + 3p + 1

So x(3) = 2p²+3p+1
And you know that x(3) = 1
so 2p² + 3p + 1 = 1 that makes p = 0 or p = -3/2
And you are said that p is different of 0 so p = -3/2
Reply 6
Avatar for paronomase
paronomase
Now show us how you would work out the value of x(2008)
Reply 7
Avatar for sammy3000
sammy3000
OP
wouldn't x2008=-0.5
because when x is odd its equal to 1 and when its even its equal to -0.5
Reply 8
Avatar for paronomase
paronomase
This sounds correct to me; however you would have to prove it.
For example, just show how you calculate x1, x2, x3 and x4 (well only x2 and x4) and then I would make an hypothesis: "For x odd...for x even..."

And then you prove it manipulating the sequence you are given.

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