Edexcel FP1 Thread - 20th May, 2016

Hi does anyone know please whether all the mad as maths questions under ellipse are in the spec? http://madasmaths.com/archive/maths_booklets/further_topics/various/conic_sections.pdf Thanks

Ellipses? You don't do ellipses in FP1, do you?

I don't think so, I think I'm paranoid haha, is it just parabolas and rectangular hyperbola that we need to be able to do for FP1? Thanks

Yep, just those two are on spec.

Hi, I was doing the June 2014 R paper and got stuck on knowing how many assumptions to make for the proof of question 9 (b). Can someone please explain the question to me?

You need only assume that:

$u_{k} = 4^{k+1} - 2^{k+3}$ and $u_{k-1} = 4^k - 2^{k+2}$.

Then: $u_{k+1} = 6u_{k} - 8u_{k-1} = \cdots$ where you substitute in from above.

Since it's a recurrence relation involving three terms, you need to make two assumptions. When there is only two terms, you make 1 assumption, etc...

Gotcha, thanks!

Hi, please could anyone explain question 4e on exercise A here https://644625398389466aee00633223056f55519d5fbc.googledrive.com/host/0B1ZiqBksUHNYQWY1UWtNa0NRNU0/CH5.pdf I can't really make sense of SB, thanks

Okay, so - first off, you can see in includes just odd numbers - in fact, it's two consecutive odd numbers multiplied together.

So, the general term is: $(2n+1)(2n+3)$, you want to add up these except starting from 3 * 5, so you need to start from $n=1$.

Hence: $\sum_{r=1}^{k} (2n+1)(2n+3) = (3)(5) + (5)(7) + \cdots+ (2k+1)(2k+3)$ and there are precisely $k$ terms in this summation, as required.

Ah I see, brilliant, thank you

Hey guys,

I am really stuck on practice paper a Qu 5 in this link:

http://www.thestudentroom.co.uk/attachment.php?attachmentid=72187&d=1244566570

Surely it is impossible to part a before part b as u can't find the real root

How can I expand this out if A and B are both matrices? I know that the order of multiplying matrices matters, so is there a particular way to know which way to order the matrices?
$\mathbf{M} = (\mathbf{A+\mathbf{B}})(2\mathbf{A-\mathbf{B}})$

Its from the June 2014 R paper, question 6.

You could just work out each bracket numerically and multiply, but if you do expand keep the order the same from left to right, so $(A + B)(2A - B) = 2A^2 +2BA - AB - B^2$