Limlove
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#1
Report Thread starter 11 years ago
#1
Hey
I've got a C2 resit on Friday and have been doing some past papers
I understand pretty much all of it but can't get my head around a certain part of the sequences chapter.

An example of a question (January 06) that i don't have a clue about is:

The nth term of a sequence is Un
The sequence is defined by U(n+1) = pUn + q

The first three terms of the sequence are given by
U1= 200 U2= 150 U3= 120

a) Show that p= 0.6 and find the value of p

I understand this and came out with p=0.6 and q= 30

b) Find the value of U4

I also understand this and came out with U4 = 102

c) The limit of Un as n tends to infinity is L. Write an equations for L and hence find the value of L

I have absolutely no clue where to start with the question, any help would be much appreciated thanks
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ghostwalker
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#2
Report 11 years ago
#2
(Original post by Limlove)
Hey
don't have a clue about is:

The nth term of a sequence is Un
The sequence is defined by U(n+1) = pUn + q

c) The limit of Un as n tends to infinity is L. Write an equations for L and hence find the value of L

I have absolutely no clue where to start with the question, any help would be much appreciated thanks
Since it says write an equation for L, you are not expected to do any calculation to find it, but here is some background you might find helpful.

OK you have:

U_{n+1}=0.6U_n+30

And you need to take the limit as n tends to infinity. So:

\displaystyle \lim_{n\to\infty}U_{n+1}=\lim_{n  \to\infty}\left(0.6U_n+30\right)

And:

\displaystyle \lim_{n\to\infty}U_{n+1}=0.6 \times \lim_{n\to\infty}U_n+30

Thus:

 L=0.6L+30
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Selkarn
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(Original post by ghostwalker)
Since it says write an equation for L, you are not expected to do any calculation to find it, but here is some background you might find helpful.

OK you have:

U_{n+1}=0.6U_n+30

And you need to take the limit as n tends to infinity. So:

\displaystyle \lim_{n\to\infty}U_{n+1}=\lim_{n  \to\infty}\left(0.6U_n+30\right)

And:

\displaystyle \lim_{n\to\infty}U_{n+1}=0.6 \times \lim_{n\to\infty}U_n+30

Thus:

 L=0.6L+30
Could be wrong, but limits? In C2? :confused:

Edit: Nevermind, different exam board
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Limlove
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#4
Report Thread starter 11 years ago
#4
(Original post by ghostwalker)
Since it says write an equation for L, you are not expected to do any calculation to find it, but here is some background you might find helpful.

OK you have:

U_{n+1}=0.6U_n+30

And you need to take the limit as n tends to infinity. So:

\displaystyle \lim_{n\to\infty}U_{n+1}=\lim_{n  \to\infty}\left(0.6U_n+30\right)

And:

\displaystyle \lim_{n\to\infty}U_{n+1}=0.6 \times \lim_{n\to\infty}U_n+30

Thus:

 L=0.6L+30


Thanks for the help ghostwalker!

I've just had a look at the mark scheme and it goes even further than that

it states the equation 'L = 0.6L +30' for one mark
but then for the the other two marks it says :

L = q / (1-p)
L = 75

I'm guessing this has something to do with the equation a/(1-r) but im not really sure where to go form there?
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Jooeee
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#5
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(Original post by Limlove)
Thanks for the help ghostwalker!

I've just had a look at the mark scheme and it goes even further than that

it states the equation 'L = 0.6L +30' for one mark
but then for the the other two marks it says :

L = q / (1-p)
L = 75

I'm guessing this has something to do with the equation a/(1-r) but im not really sure where to go form there?
It's just rearranged to find the limit.

You may find it easier to do that this way though:

 L=0.6L+30



L-0.6L=30



0.4L=30



L=30/0.4



L=75
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ghostwalker
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#6
Report 11 years ago
#6
(Original post by Limlove)
Thanks for the help ghostwalker!

I've just had a look at the mark scheme and it goes even further than that

it states the equation 'L = 0.6L +30' for one mark
but then for the the other two marks it says :

L = q / (1-p)
L = 75

I'm guessing this has something to do with the equation a/(1-r) but im not really sure where to go form there?
As the question says, state the equation.

Then you need to solve the equation, to find L.
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Limlove
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(Original post by Jooeee)
It's just rearranged to find the limit.

You may find it easier to do this though:

 L=0.6L+30



L-0.6L=30



0.4L=30



L=30/0.4



L=75
Thanks joe sooooo much easier, can't believe i didnt spot it! lol
TSR saves me again!!
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M.Johnson2111
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#8
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#8
(Original post by Limlove)
Hey
I've got a C2 resit on Friday and have been doing some past papers
I understand pretty much all of it but can't get my head around a certain part of the sequences chapter.

An example of a question (January 06) that i don't have a clue about is:

The nth term of a sequence is Un
The sequence is defined by U(n+1) = pUn + q

The first three terms of the sequence are given by
U1= 200 U2= 150 U3= 120

a) Show that p= 0.6 and find the value of p

I understand this and came out with p=0.6 and q= 30

b) Find the value of U4

I also understand this and came out with U4 = 102

c) The limit of Un as n tends to infinity is L. Write an equations for L and hence find the value of L

I have absolutely no clue where to start with the question, any help would be much appreciated thanks
Hi, how did you get U4 for part b?
Just sat this paper for revision
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Plücker
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(Original post by M.Johnson2111)
Hi, how did you get U4 for part b?
Just sat this paper for revision
By now Limlove has probably forgotten that TSR exists.
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M.Johnson2111
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(Original post by Plücker)
By now Limlove has probably forgotten that TSR exists.
Probably but hopefully not, the mark sheme doesnt provide much help and I can't find other online solutions
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Plücker
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(Original post by M.Johnson2111)
Probably but hopefully not, the mark sheme doesnt provide much help and I can't find other online solutions
If you did part a then you know p and q.

u_4 = p u_3+ q
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M.Johnson2111
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Thank you, your format makes a lot more sense where it says U3 I was writing U5 so must of wrote n+1 anyhow thank you
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