Maths C2 help please

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

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, ...

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 ?

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

3logx and log x^3 are the same thing so either is fine.

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.

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?

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

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

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

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 ...

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.

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.