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C1 Maths extra practise

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4+ 2root6 + 6root2 + 2root3
Original post by Zacken
On the mean side, but whatever - show that:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\frac{\sqrt{3} + \sqrt{2} - 1}{\sqrt{3} - \sqrt{6} + 1} = 2 + a\sqrt{2} + b\sqrt{3} + c\sqrt{6}\end{equation*}



where a,b,ca,b,c are integers to be determined.

Spoiler

(edited 7 years ago)
Original post by not_lucas1
4+ 2root6 + 6root2 + 2root3

You're supposed to get 2root3-2
Try this for the denominator: (root3(1-root2)+1)(root3(1+root2)+1) which gives 2root3-2. It's easier because it's in quadratic form. Then for the numerator do this (root3 + root2 -1)(root3(1+root2)+1)

Basically factories the denominator to make it easier p.s.(quote me so I can see the notification)
(edited 7 years ago)
Original post by saffronlord
isnt this c2?


You might be right :s-smilie: I do not know.
Reply 44
Original post by SeanFM
Here is a question for you to try :h:


A sequence of numbers a0,a1,a2,... a_0, a_1, a_2, ... is given by an+1=2an+64n a_{n+1} = 2a_{n} + 6 - 4n.

Given that a2=6a_2 = 6, find a1a_1 and a3a_3.

Hence find the value of i=110ai\sum_{i=1}^{10} a_{i}.

Hence, or otherwise, find the value of i=010ai\sum_{i=0}^{10} a_{i}.


Would they ever ask you that?
Reply 45
Original post by YeSand
Would they ever ask you that?


the question is standard, the last question would only be worth 1 mark I think.
Reply 46
Original post by imran_
the question is standard, the last question would only be worth 1 mark I think.


For this:
Hence find the value of .


You need to do find a1 - 10 and add them up.

That will take a lot of time in the exam. They will never ask you something this ridiculous. Going up to a6 is possibly the furthest they should go.
Reply 47
Original post by imran_
the question is standard, the last question would only be worth 1 mark I think.


The last question you find a0 and add it to the previous a1-10. Simple.
Original post by YeSand
Would they ever ask you that?


I don't think it'd be as complicated as that. Plus, some of it might be some C2 stuff.
Original post by YeSand
For this:
Hence find the value of .


You need to do find a1 - 10 and add them up.

That will take a lot of time in the exam. They will never ask you something this ridiculous. Going up to a6 is possibly the furthest they should go.


you just use the sequence equation Sn=n/2(2a+(n-1)d) or Sn=n/2(a+l)
(edited 7 years ago)
Original post by EndOfTheTour
Has anyone got access to the C1 mymaths? Is it worth using? My school's password doesn't seem to be working, should I even bother?


I find MyMaths rubbish for explaining concepts for A level. The best revision tool is Exam Solutions.
Reply 51
c1 has never been hard, hopefully they dont make it hard this year
Reply 52
Original post by XOR_
you just use the sequence equation Sn=n/2(2a+(n-1)d) or Sn=n/2(a+l)


That's a common error. Those formulas won't work on recurrent sequences.
Reply 53
Original post by surina16

Spoiler



Original post by not_lucas1
cant solve it atm but multiply the whole thing by the denominator with a flipped sign so - to + then simplify it? is that correct?
Original post by Someboady
Er are these the correct answers:I probably messed up somewhere :/

Spoiler

Original post by odera_dim
a=1b=1c=1This was a nasty question and I'm not too sure about the answer but this is what I got 😩
Original post by techfan42
Same, I got that too
Original post by JN17
^ I also got a,b,c all equal to 1
Original post by imran_
this question is taking me too long smh



Unparseable latex formula:

\displaystyle[br]\begin{equation*}\frac{\sqrt{3} + \sqrt{2} - 1}{1 + \sqrt{3} - \sqrt{6}} \times \frac{1 + \sqrt{3} + \sqrt{6}}{1 + \sqrt{3} + \sqrt{6}} = \frac{1 + 2\sqrt{2} + \sqrt{3}}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3}+1} = 2 + \sqrt{2} + \sqrt{3} + \sqrt{6}\end{equation*}

Reply 54
Original post by YeSand
That's a common error. Those formulas won't work on recurrent sequences.


thats why you need to work them out
after working out a1 that gives you the first term a=2
work out a2 and a3 you'll get the next terms a2=6 a3=10

then work out the difference and that would be your (d) d=4

N= 10 cos theres 10 terms

now work from there
Original post by Zacken
Unparseable latex formula:

\displaystyle[br]\begin{equation*}\frac{\sqrt{3} + \sqrt{2} - 1}{1 + \sqrt{3} - \sqrt{6}} \times \frac{1 + \sqrt{3} + \sqrt{6}}{1 + \sqrt{3} + \sqrt{6}} = \frac{1 + 2\sqrt{2} + \sqrt{3}}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3}+1} = 2 + \sqrt{2} + \sqrt{3} + \sqrt{6}\end{equation*}



eh I was only 5 root 2s off :biggrin:

I was lost at the first step because of the 2 surds on the bottom and stupidly multiplied by (root6 - root3 + 1) :colondollar:

Spoiler

Reply 56
Original post by surina16
eh I was only 5 root 2s off :biggrin:

I was lost at the first step because of the 2 surds on the bottom and stupidly multiplied by (root6 - root3 + 1) :colondollar:

Spoiler



Yes, that's the tricky bit. :smile:

Stuff like this came up in my Year 11 IGCSE all the time. :tongue:
Reply 57
Original post by imran_
thats why you need to work them out
after working out a1 that gives you the first term a=2
work out a2 and a3 you'll get the next terms a2=6 a3=10

then work out the difference and that would be your (d) d=4

N= 10 cos theres 10 terms

now work from there


Be careful. Recurrence sequences do not have same common difference.
Original post by Zacken
Yes, that's the tricky bit. :smile:

Stuff like this came up in my Year 11 IGCSE all the time. :tongue:


:getmecoat:
Fam C1 is easy n sounds like you already done enough revision

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