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# Reccurence relations

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1. https://a9497d7f220e174d38d014b2b1d8...%20Edexcel.pdf

Question 3

how do i know that =1?
2. (Original post by thefatone)
https://a9497d7f220e174d38d014b2b1d8...%20Edexcel.pdf

Question 3

how do i know that =1?
You don't. You know that
3. (Original post by nerak99)
You don't. You know that
but you do to reach that conclusion

look you just put in n=1 to get out of the reccurence relation what you've stated but how do i know that

n=1???
4. Well, for a start, I would not really call this a recurrence relation. It is a formula where we are told that
Hence, if n=1 .

Otherwise, I do not understand your question. Yes, I just put in n=1 but that is what is required here.

It turns out that anyway.
5. (Original post by nerak99)
Well, for a start, I would not really call this a recurrence relation. It is a formula where we are told that
Hence, if n=1 .

Otherwise, I do not understand your question. Yes, I just put in n=1 but that is what is required here.
Here's how I'd do it.

as we know u1 = u3, you can just put in the algebra:

2^1 + k = 2^3 +3k then it's simple rearranging

2^1 = 2^3 + 2k
0 = 6 + 2k
-6 = 2k
-3 = k
6. (Original post by nerak99)
Well, for a start, I would not really call this a recurrence relation. It is a formula where we are told that
Hence, if n=1 .

Otherwise, I do not understand your question. Yes, I just put in n=1 but that is what is required here.
how do you know this where have you gotten n=1 from? why isn't n=2?
7. We are told in a) that and we have used that fact.

I think you should go off and have a rest and come back later.

From your previous post have you not done C1 or was that agonising just about a practice paper?
8. (Original post by thefatone)
how do you know this where have you gotten n=1 from? why isn't n=2?
The n is different for each term. In the case of u1 n=1
in the case of u3 n = 3
9. (Original post by nerak99)
We are told in a) that and we have used that fact.

I think you should go off and have a rest and come back later.

From your previous post have you not done C1 or was that agonising just about a practice paper?
that was me being super annoyed about the fact that i can't not make stupid mistakes
yes i do need a rest but i can't afford one, i need to do more C1
(Original post by Malandirix)
The n is different for each term. In the case of u1 n=1
in the case of u3 n = 3
i see so the number below the u is the number n= ..... right thanks
10. Well that is OK. I will take your future rep award as an apology.
11. (Original post by thefatone)
...
The formula says that the given relation is true for all natural n. You can plug in whatever n you want, plug in n=2 if you want. It'll just be utterly useless and a waste of your time. The only useful ones you get is from and . If you can't see why that is the most useful ones to plug in, then you really do need to have a rest.
12. (Original post by Zacken)
The formula says that the given relation is true for all natural n. You can plug in whatever n you want, plug in n=2 if you want. It'll just be utterly useless and a waste of your time. The only useful ones you get is from and . If you can't see why that is the most useful ones to plug in, then you really do need to have a rest.
i guess but it can't be helped, i've tanned out these C1 and i need to do more, i can't do less, it won't help(yes yes overworking will exhaust me etc and i may not learn anything) but then again any new format of question i come across and i do i will remember and know how to do it the next time....

uberteknik looks like your advice was good ^-^ and other ppl say the same thing too xD
13. (Original post by thefatone)
i guess but it can't be helped, i've tanned out these C1 and i need to do more, i can't do less, it won't help(yes yes overworking will exhaust me etc and i may not learn anything) but then again any new format of question i come across and i do i will remember and know how to do it the next time....

uberteknik looks like your advice was good ^-^ and other ppl say the same thing too xD
Okay.
14. (Original post by Zacken)
Okay.

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