Edexcel AS Mathematics C2- 25th May 2016 Thread

I found this binomial expansion question on the June 2014 IAL paper:

Given that 1 + 12x + 70x² + …

is the binomial expansion, in ascending powers of x of (1 + bx)ⁿ, where n (*insert symbols*)…and b is a constant,

(a) show that nb = 12 [ 1 mark ]

(b) find the values of the constants b and n. [ 6 marks ]

Do you think something like this may be in the exam on Wednesday?

It could be asked.

You do the same thing for the $x^{2}$ term and then solve simultaneously

Edit: It's probably worth noting that IAL papers are a little bit tricker than UK papers, this is represented in questions like these

E.g. the c1 paper the other day had the same type of discrimimant question on jan 2016 ial

someone post the JAN 2016 IAL c12 paper please, I cant find it anymore

I found this binomial expansion question from the June 2014 IAL paper:

Given that 1 + 12x + 70x² + …

is the binomial expansion, in ascending powers of x of (1 + bx)ⁿ, where n (*insert symbols*)…and b is a constant,

(a) show that nb = 12 [ 1 mark ]

(b) find the values of the constants b and n. [ 6 marks ]

Do you think something like this may be in the exam on Wednesday?

I don't think it will be asked. In the exam they either ask you to find the first three, four terms of the expansion, and could either ask you give the estimation for instance.

do madasmaths papers, you'll feel more prepared.
These papers are designed to test every every part of the spec in depth. Theyre good practise

Try IYGB papers S and T, if you can complete them, you can do anything

Hi fellow AS kids (who have exams other than Maths this week), how are you studying for C2? Anyone just gonna flip through the textbook on the morning of the exam?

With this question, I don't understand how the geometric progression forms? I mean with ((1250 x 1.06) + 1250) x 1.06, when 1250 is factorised out isn't the +1250 missed out??

At the end of the first year we have $U_1 = ( 1250 )( 1.06 )$, at the end of the second year we have $U_2 = ( U_1 + 1250 )( 1.06 )$ etc...

Simplifying: $U_2 = (1250(1.06) + 1250)(1.06) = 1250(1.06)^{2} + 1250(1.06)$

Now at the start of a year we add 1250, so for the nth year, the value at the start is $V_{n} = U_{n-1} + 1250$

So for the third year it's:

$V_{3} = U_{2} + 1250 = 1250(1.06)^{2} + 1250(1.06) + 1250 = \displaystyle \sum_{i=1}^{n} 1250(1.06)^{n-1}$

Thank you I think I tried to add all the individual terms as well, but that makes more sense

I finished all the normal past papers, but going to do the IALs and solomon now

- x8 C2 exam past papers = ---->
^ 8 x 1.5h = 12 hours ---> all day I guess the faster I complete them the longer breaks I'll have at the end.
9/9 C2 past papers to be completed for Wednesday exam.

Be careful to not overdo it, the killer in C2 is usually silly mistakes rather than lack of knowledge

Advice for the exam: Take it at quite a quick but accurate pace, make sure you finish ahead of time and check literally every single thing ($3 + 4 \neq 5$ etc..)

Good luck

Thanks bud! We're all gonna make it! (i hope)