FP1 proof

What I would have a problem with is that you've not shown me that $n^3 + 2n^2 + 7n + 14$ factorises to $(n+2)(n^2 + 7)$. I would say you would need to show how you have factorised this, but using the formula to get to this stage is fine.

I am asked to prove the following:


Am I simply allowed to do this step after using the summation formulae and conclude that it is true?


... or does the examiner expect me to include the intermediate steps?

If your summation formula is in your formula book, I think this is fine. The examiner will know that you have just used the formula from your book and taken this legal short cut (as opposed to just stating it from memory).

What I would have a problem with is that you've not shown me that $n^3 + 2n^2 + 7n + 14$ factorises to $(n+2)(n^2 + 7)$. It looks like you've just assumed it because the question says it, its not really an easy factorisation to notice lol. I would say you would need to show how you have factorised this, but using the formula to get to this stage is fine.

Since it's a proof, you should really include all the steps you take.

$\frac{1}{4}n[n^3 + 2n^2 + 7n + 14] = \frac{1}{4}n[n(n^2+7) + 2(n^2 +7)]$

Yeah that's what I was worried about. I'm not too confident when trying to factorise cubics, so would expanding out the step I am trying to get to and comparing with my unfactorised bracket suffice?

Would it be fine to look at the given expression, looking for linear factors, dividing getting the quadratic and then factor that from there?

Seeing as your doing FP1, I assume you know how to factorise cubics yeah?

Would it be fine to look at the given expression, looking for linear factors, dividing getting the quadratic and then factor that from there?

Nope. Not part of my course (Edexcel) I don't believe. I'll learn how to now then.

To factorise? Yeah, thats fine. You've shown that you've not just guessed what the answer is, but how to get to the answer.

Thanks.
While im at it, if you get a question on the sum from r = 1 to 2n like $\sum\limits_{r=1}^{2n} r$ would it simply be the sum to r = 1 to n but with 2n instead of n? If so why?

Is factor theorem not in C2?
If not you'll certainly come across it in C3

That would all be a bit unnecessary, but you could do it like that.

In FP1 proofs by induction, you often need to simplify complex expressions, but all of this can easily be done without use of the factor theorem/long division.

Use the formulae in the data booklet and replace n with 2n



Yes factor theorem is in C2, but at a very basic level where they always give you a solution, unlike in this.

And in C1 cubics show up but you can always take a factor of x outside.

Nope. Not part of my course (Edexcel) I don't believe. I'll learn how to now then.

Easiest way I do it is:

1) By trial and error, you find a solution such that your function is 0. So let's say you have some function, f(x). You want to find a value for x, such that f(x) = 0.

This is normally just done by guessing, but they tend to make the numbers easy for this. So you would really just try like x = ±1, ±2, ±3 and so on. You wouldn't need to go too far until you find your solution.

2) Let's say that at some point 'a', you get f(a) = 0. What this means is that x = a is solution to your function (i.e a point where it crosses the 'x' axis) and so we can say that (x - a) = 0 is a root. With your example, notice how n = -2 is a solution and so n + 2 = 0 is a root.

3) Now you have two options. You can either use the long division way and divide f(x) by (x - a) and this will give you a quadratic. Either that or you can compare coefficients but i don't use this method so don't know how to explain it. You can then factorise this quadratic in the normal way.

If you tried dividing your cubic by n + 2, you should get n² + 7. This can't be factorised any more so your final answer is what you got.

Another way to do it is the way Indeterminate has done it.

You have a cubic which is

$n^3 + 2n^2 +7n + 14$.

What we see here is that 3 terms contain an 'n' and two terms contain a '2'. So lets take the two terms containing a '2' and then take the other two terms left that have an 'n' and so we can re-write and factorise to get

$n^3 + 7n + 2n^2 + 14 = n(n^2 + 7) + 2(n^2 + 7)$.

Now clearly, we have one common factor here and so we can factorise this out and get

$(n^2 + 7)(n + 2)$.

In edexcel fp1 when proving summation formula, it is proved by mathematical induction? Show that LHS = RHS is true for n=1 then assumption n = k followed by n = k + 1

Im pretty sure there's no such shortcut to proofs in fp1? Have you learnt this?

I think other replies in the thread may have done a different exam board

This could be an induction question or it could be a proof using standard results. The question will clearly state if induction is required.