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Edexcel FP1 Thread - 20th May, 2016

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Original post by WhatIsSleep
I'm stuck on this question (part b), I can't seem to show it's divisible by 8 and the mark scheme's no help :/


make the 4(2^k) divisible by 8 by doing 4*2*(2^k-1) which equals 8^(2^k-1)
Original post by Music With Rocks
Sounds doable (but difficult of course)

Yep, this year I am doing C3, C4, M2, S2, S3, FP2 (+FP1 resit)
Last year I did C1, C2, D1, S1, M1, FP1


Nice - good luck with them all! c3 will be the module to target for the A* maths which is the most important thing
Original post by maruchan
make the 4(2^k) divisible by 8 by doing 4*2*(2^k-1) which equals 8^(2^k-1)


Yes haha, it didn't cross my mind for some reason
Thank you very much!
Original post by WhatIsSleep
Yes haha, it didn't cross my mind for some reason
Thank you very much!


That's ok good luck with your exam!
Original post by thesmallman
Question guys: In a summation series formula, when n comes up as well as r
e.g. sum of (0 to n): 2r^(3) + r^(2) + 2n + 1
Is 2n+1 a constant? i.e. in sum of (0 to 10)- you would convert the first two terms into 'n' in the normal way then sub in '10'.
Then, for the last term you'd convert it into n(2n+1) and also sub in 10??

Thanks!


you done it?
So what's everyone's plan for tonight- go to bed early or all nighter? :biggrin:
Original post by tazza ma razza
you done it?


yeahhh I've got it, thanks!
Original post by thesmallman
yeahhh I've got it, thanks!


awesome.
Original post by tazza ma razza
Nice - good luck with them all! c3 will be the module to target for the A* maths which is the most important thing


Good luck to you too with all your exams! Yeah will definitely have to put a lot of work into the cores for that A*
Original post by tazza ma razza
nicely done mate! tbh im expecting 90 ums minimum tomorrow - should be alright lol

i read it as s * r^2 -> when it should be read as (SP)^2. is that right hence why we wouldnt root it


(SR)2(SR)^2, yes, from Pythagoras.
Original post by NotNotBatman
(SR)2(SR)^2, yes, from Pythagoras.


you da man! (Y)
Reply 631
Avatar for Eshanth
Eshanth
Original post by WhatIsSleep
I'm stuck on this question (part b), I can't seem to show it's divisible by 8 and the mark scheme's no help :/


What paper is that? I got an answer but i'm not sure if it's right.
Original post by Eshanth
What paper is that? I got an answer but i'm not sure if it's right.


June 2010 Q7
Good luck for tomorrow guys! I'm dreading it hahaha
Original post by Nikhilm
In this case you would prove n = 0 (instead of n = 1) right?


Yes, it'd be the same as normal induction but you just start with n=0.
Reply 635
Avatar for Eshanth
Eshanth
Original post by WhatIsSleep
June 2010 Q7


Ok, if you got part a right, part b is easy because for f(k+1), you can use part a answer, then from that step you know that 6f(k) is divisible by 8 but the second part isn't so you have to get 8 for the -4(2^k) so what you do is simply take a 2 outside of the 2^k which then becomes 2^k-1 and the -4 at the front muliplies by 2 making it -4x2(2^k-1) which is -8(2^k-1) which is divisible by 8! Do you get it?
Original post by Eshanth
Ok, if you got part a right, part b is easy because for f(k+1), you can use part a answer, then from that step you know that 6f(k) is divisible by 8 but the second part isn't so you have to get 8 for the -4(2^k) so what you do is simply take a 2 outside of the 2^k which then becomes 2^k-1 and the -4 at the front muliplies by 2 making it -4x2(2^k-1) which is -8(2^k-1) which is divisible by 8! Do you get it?


Yes (: Thank you for explaining it so well
Original post by SB0073
Ahh want to make it 8/9 Pleeeease! Just incase :smile: I remember seeing this in a past paper last question just cant find it anymore! :O Didn't understand it fully at the time

If I had to predict something coming up I think a proof by induction of Un+2=... type of question may come up!


Read this for the question, and then here's my answer:

Two ways to look at it:

1. ∑r=0n(2n+1)=2(n)+1⏟0th   term+∑r=1n(2n+1)=(2n+1)+(2n+1)n=(2n+1)(n+1)\displaystyle \sum_{r=0}^n (2n+1) = \underbrace{2(n) + 1}_{0\text{th \, term}} + \sum_{r=1}^n (2n+1) = (2n+1) + (2n+1)n = (2n+1)(n+1)2.

2. ∑r=0nf(r)=f(0)+f(1)+f(2)+⋯+f(n)⏟n terms⏞(n+1) terms\displaystyle \sum_{r=0}^n f(r) = \overbrace{f(0) + \underbrace{f(1) + f(2) + \cdots +f(n)}_{n \, \text{terms}}}^{(n+1) \, \text{terms}}.

Since in this case you have ∑r=0n(2n+1)=(2n+1)∑r=0n1\sum_{r=0}^n (2n+1) = (2n+1)\sum_{r=0}^n 1Then you have f(r)=1f(r) = 1 being added together n+1n+1 times to get a total of (2n+1)(1+1+⋯+1)=(2n+1)(n+1)(2n+1)(1 + 1 + \cdots +1) = (2n+1)(n+1).


Original post by Nikhilm
Mathematical induction Q:

say if you have a summation formula, 1 to n, for (2n + r^2 + 4r). I.e. the n is within the equation. It would be impossible to prove an equality by mathematical induction right? Because the actual equation would change when you have k or k+1?

@Zacken


It's certainly possible to prove it using induction, but it'd be a pain in the ass and it's the type of sum that's you'd "prove" using the the standard summation formula given in the formula booklet by splitting and then simplifying. Look above for knowing how to sum n. Remember it's just a fixed unchanging constant.

Original post by NotNotBatman
So, matrix multiplication is not commutative, but for the induction questions it is, because (where A\mathbf A is a matrix)

Unparseable latex formula:

\mathbf{A}^k^+^1 = \mathbf{A}^k\cdot \mathbf{A}^1

and
Unparseable latex formula:

\mathbf{A}^k^+^1 = \mathbf{A}^1\cdot \mathbf{A}^k

due to index laws, so does it matter what way around we multiply it in the in inductive proof?


It's not commutative in general in the sense that AB≠BAAB \neq BA for two different matrices A,BA,B but it is associative so Ak+1=AAk=AkAA^{k+1} = AA^{k} = A^{k}A, so it's all cool in induction questions.

Original post by NotNotBatman
For divisibility tests, is the conclusion different from true for n=k, k+1,n =1, so all n. Do you have to state since f(k) divisible by n, so f(k+1) is divisible by n etc.


Well, it'd be better to say "since f(k) divisible by n..." but either is fine.
(edited 7 years ago)
Original post by Zacken
I'll give you two ways to look at it:

1. ∑r=0n(2n+1)=2(n)+1⏟0th   term+∑r=1n(2n+1)=(2n+1)+(2n+1)n=(2n+1)(n+1)\displaystyle \sum_{r=0}^n (2n+1) = \underbrace{2(n) + 1}_{0\text{th \, term}} + \sum_{r=1}^n (2n+1) = (2n+1) + (2n+1)n = (2n+1)(n+1)

2. ∑r=0nf(r)=f(0)+f(1)+f(2)+⋯+f(n)⏟n terms⏞(n+1) terms\displaystyle \sum_{r=0}^n f(r) = \overbrace{f(0) + \underbrace{f(1) + f(2) + \cdots +f(n)}_{n \, \text{terms}}}^{(n+1) \, \text{terms}}.

Since in this case you have ∑r=0n(2n+1)=(2n+1)∑r=0n1\sum_{r=0}^n (2n+1) = (2n+1)\sum_{r=0}^n 1

Then you have f(r)=1f(r) = 1 being added together n+1n+1 times to get a total of (2n+1)(1+1+⋯+1)=(2n+1)(n+1)(2n+1)(1 + 1 + \cdots +1) = (2n+1)(n+1).


Yup, first way is best IMO.
I think simultaneous equations w/ matricies will come up - never seen it before and i have no idea how to do them lol

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