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FP1 Proof

Hi I was wondering if someone could help me answer edexcel January 2011 question 9 proof by induction. I don't understand the first part when you have to rearrange it could someone help me?



https://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/January%202011%20-%20QP/6667_01_que_20110131.pdf
Original post by Sapphiresmith
Hi I was wondering if someone could help me answer edexcel January 2011 question 9 proof by induction. I don't understand the first part when you have to rearrange it could someone help me?



https://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/January%202011%20-%20QP/6667_01_que_20110131.pdf

Sorry, I don't understand "I don't understand the first part when you have to rearrange it". What do you have so far?
Original post by Smaug123
Sorry, I don't understand "I don't understand the first part when you have to rearrange it". What do you have so far?



Nothing I don't understand the first part when I have to rearrange un+1 into un
Original post by Sapphiresmith
Nothing I don't understand the first part when I have to rearrange un+1 into un

But un+1u_{n+1} is already given as a function of unu_n. No rearranging is needed to put un+1u_{n+1} in terms of unu_n.
Original post by Smaug123
But un+1u_{n+1} is already given as a function of unu_n. No rearranging is needed to put un+1u_{n+1} in terms of unu_n.


In the mark scheme it does and I don't understand how they have done it

https://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/January%202011%20-%20MS/6667_01_rms_20110309.pdf
Original post by Sapphiresmith

Ah, if you'd said that you were trying to follow the worked solution and given a link to it, that would have helped.

I still don't see where they "rearrange un+1u_{n+1} into unu_n". Is there a specific line of the solution you are unhappy with?
Original post by Smaug123
Ah, if you'd said that you were trying to follow the worked solution and given a link to it, that would have helped.

I still don't see where they "rearrange un+1u_{n+1} into unu_n". Is there a specific line of the solution you are unhappy with?



Sorry that was my mistake I should have posted it in the first post!

But if you look at the first line it forms another equation of from the first one which I am unsure of
Original post by Sapphiresmith
Sorry that was my mistake I should have posted it in the first post!

But if you look at the first line it forms another equation of from the first one which I am unsure of

The first line is just their restating the problem. The first line of my solution might look like:
un=defn4un−1+2=?23(4n−1)\displaystyle u_{n} \overset{\text{def}^\text{n}}{=} 4 u_{n-1} + 2 \overset{?}{=} \frac{2}{3} (4^n - 1)

which is the same but with some things written above equals signs.
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Original post by Sapphiresmith
Hi I was wondering if someone could help me answer edexcel January 2011 question 9 proof by induction. I don't understand the first part when you have to rearrange it could someone help me?



https://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/January%202011%20-%20QP/6667_01_que_20110131.pdf



I'm not too sure by what you're confused of.

We have the recurrence relation un+1=4un+2u_{n+1} = 4u_{n} + 2. The question wants you to prove that the general series for the recurrence relation is given by un=23(4n−1)\displaystyle u_n = \frac{2}{3}(4^n - 1). The nth term of the recurrence relation, if you will.

So we prove that the u1=2u_1 = 2 using u1=23(4−1)u_1 = \frac{2}{3}(4-1).

Then we assume the result is true for n=kn=k, i.e: uk=23(4uk−1)u_k = \frac{2}{3}(4u_k -1).

Now we need to prove this for n=k+1n=k+1. We know that n=k+1n=k+1'th term is given by:

uk+1=4uk+2u_{k+1} = 4u_k + 2, then you substitute in for uku_k and prove the result.
(edited 9 years ago)

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