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# Maths C2 help please watch

1. a couple quick questions!;

what's the nth term for this sequence? I'm struggling as there's (I don't think) no common ratio (and so can't use the formula ar^(n-1) and it's surely not an AP either?
4, 9, 16, 25, ...

and a bit stuck with this one;

how do I simplfy (into a single log);
logx+logx^2

just unsure if I do
logx+2logx = 3logx
or
times the x^2 by the x, and so =logx^3 ?

many thanks
2. (Original post by m1800)
a couple quick questions!;

what's the nth term for this sequence? I'm struggling as there's (I don't think) no common ratio (and so can't use the formula ar(n-1) and it's surely not an AP either?
4, 9, 16, 25, ...

and a bit stuck with this one;

how do I simplfy (into a single log);
logx+logx^2

just unsure if I do
logx+2logx = 3logx
or
times the x^2 by the x, and so =logx^3 ?

many thanks
No it's not an AP. Do you recognise these numbers?

3logx and log x^3 are the same thing so either is fine.
3. (Original post by m1800)
what's the nth term for this sequence? I'm struggling as there's (I don't think) no common ratio (and so can't use the formula ar(n-1) and it's surely not an AP either?
4, 9, 16, 25, ...
You are correct that this is not a geometric or arithmetic sequence. You are expected to notice something - these are numbers that you should recognise. If I tell you that the term before the 4 is 1, and the term after the 25 is 36, does that help?

(Original post by m1800)
how do I simplfy (into a single log);
logx+logx^2

just unsure if I do
logx+2logx = 3logx
or
times the x^2 by the x, and so =logx^3 ?
It looks like you have used the relationship log(x^n) = n log(x) to turn log(x^2) into 2 log (x), so you can use this same law to see that your two suggestions are equivalent. If you are asking which form they will be expecting, I don't think it really matters. Although there might be a case for saying that log(x^3) is a single logarithm, rather than three times a single logarithm, but I think that's a bit overly pedantic.
4. For the sequence, try taking the differences between each term and then finding the second-order differences (the differences of the differences). Those should be constant. Do you remember doing quadratic sequences at GCSE?

As for the log, remember that log(x^2) (I'm assuming that's what you meant) is 2logx.
5. (Original post by Notnek)
No it's not an AP. Do you recognise these numbers?

3logx and log x^3 are the same thing so either is fine.
(Original post by Pangol)
It looks like you have used the relationship log(x^n) = n log(x) to turn log(x^2) into 2 log (x), so you can use this same law to see that your two suggestions are equivalent. If you are asking which form they will be expecting, I don't think it really matters. Although there might be a case for saying that log(x^3) is a single logarithm, rather than three times a single logarithm, but I think that's a bit overly pedantic.

oh so it's correct? thanks guys.

(Original post by Pangol)
You are correct that this is not a geometric or arithmetic sequence. You are expected to notice something - these are numbers that you should recognise. If I tell you that the term before the 4 is 1, and the term after the 25 is 36, does that help?

(Original post by Notnek)
No it's not an AP. Do you recognise these numbers?

(Original post by TheMindGarage)
For the sequence, try taking the differences between each term and then finding the second-order differences (the differences of the differences). Those should be constant. Do you remember doing quadratic sequences at GCSE?
still lost :l

So the difference is of course increasing by two each time, i just have no idea how to put this into a formula :l
6. (Original post by m1800)
still lost :l

So the difference is of course increasing by two each time, i just have no idea how to put this into a formula :l
The formula for the nth term is in the format an^2 + bn + c

The second-order difference is 2, so a is equal to half of that or 1 (don't ask why...).

So now you have n^2 + bn + c. Substitute in two values (for example, n=1 and n=2):

1+b+c = 4
4+2b+c = 9

Now just solve as a normal pair of simultaneous equations.
7. (Original post by TheMindGarage)
The formula for the nth term is in the format an^2 + bn + c

The second-order difference is 2, so a is equal to half of that or 1 (don't ask why...).

So now you have n^2 + bn + c. Substitute in two values (for example, n=1 and n=2):

1+b+c = 4
4+2b+c = 9

Now just solve as a normal pair of simultaneous equations.
ok so b=2 and c=1 so whats the formula? :l

sorry ive never come across this method inc. at GCSE

to me this just seems like a very odd question altogether
8. (Original post by m1800)
ok so b=2 and c=1

sorry ive never come across this method inc. at GCSE

to me this just seems like a very odd question altogether.
This may be a valid method (I haven't read the details), I am sure that the idea of this question is to just notice a pattern, one that should be very familiar to you. 1 4 9 16 25 36 49 ...
9. (Original post by Pangol)
This may be a valid method (I haven't read the details), I am sure that the idea of this question is to just notice a pattern, one that should be very familiar to you. 1 4 9 16 25 36 49 ...
oh lord help me

of course it's square numbers

I need to sleep

thanks a lot guys
10. (Original post by Pangol)
This may be a valid method (I haven't read the details), I am sure that the idea of this question is to just notice a pattern, one that should be very familiar to you. 1 4 9 16 25 36 49 ...
You could do that, and then you'd get (n-1)^2 since the actual sequence starts with 4 rather than 1.
11. (Original post by TheMindGarage)
You could do that, and then you'd get (n-1)^2 since the actual sequence starts with 4 rather than 1.
Yes, I know that. This was my second post on this thread gently nudging the OP towards noticing the pattern, and I had already indicated that I was adding a term on the beginning as well as a few on the end.
12. (Original post by Pangol)
Although there might be a case for saying that log(x^3) is a single logarithm, rather than three times a single logarithm, but I think that's a bit overly pedantic.
If I had to choose, definitely 3 log x beats log x^3. Easier to evaluate, easier expression to differentiate/integrate, etc...

[Not that it really matters, of course! ]

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