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Fp1 proof by induction

Anyone mind checking this for me. Probably my weakest area. Cheers

http://m.imgur.com/pLaidbh,UoNfZBb
I'm seeing no problems with that. Nice work..
Where you have "True for n=1" I would add "since LHS = RHS" to make it a little clearer.
Plus I believe it is assume true for n=k

Only minor things.

That's for Edexcel - you look like u're doing OCR?
Reply 2
Original post by kavindajd
I'm seeing no problems with that. Nice work..
Where you have "True for n=1" I would add "since LHS = RHS" to make it a little clearer.
Plus I believe it is assume true for n=k

Only minor things.

That's for Edexcel - you look like u're doing OCR?

Yh on OCR. I think it may be assume true for n=k. You know the line under asume true for n=k+1, can I write that because technically that is what I am trying to show?
Reply 3
Original post by Super199
Anyone mind checking this for me. Probably my weakest area. Cheers

http://m.imgur.com/pLaidbh,UoNfZBb


Mostly fine. You want to say "assume true for n=k" and then show "n=k+1" is true. Not "assume true for n=k+1". But the rest is all fine.
Reply 4
Original post by Super199
Yh on OCR. I think it may be assume true for n=k. You know the line under asume true for n=k+1, can I write that because technically that is what I am trying to show?


Write it with a small question mark over the equals sign:

Ak+1=?A^{k+1} \stackrel{?}{=} \cdots
Original post by Super199
Yh on OCR. I think it may be assume true for n=k. You know the line under asume true for n=k+1, can I write that because technically that is what I am trying to show?


Its "With n=k+1, the matrix equation becomes:"
Reply 6
Original post by Zacken
Write it with a small question mark over the equals sign:

Ak+1=?A^{k+1} \stackrel{?}{=} \cdots


What in the actual proof?

Do you kind clarifying 8iii.
Ive done 8i and 8ii
The values that are singular is a=0 and a=2
So for iii I thought there would be solutions for a=3 but none for a=2. Whats the examiners report saying?

http://m.imgur.com/a4JJx4n,4sDg9Qh
Reply 7
Original post by Super199
What in the actual proof?


No, just in that one line where you write down what you're supposed to be showing but you haven't actually shown it yet.

For the matrix, try actually writing down the equations with the values of a subbed in, you'll see equation one is a multiple of equation 3.
Reply 8
Original post by Zacken
No, just in that one line where you write down what you're supposed to be showing but you haven't actually shown it yet.

For the matrix, try actually writing down the equations with the values of a subbed in, you'll see equation one is a multiple of equation 3.

What does that show?
Reply 9
Original post by Super199
What does that show?


That they are the same line, hence they have solutions.
Reply 10
Original post by Zacken
That they are the same line, hence they have solutions.


http://m.imgur.com/srgzWGU,J2u5gTV

9iii. Help me with the algebra pls haha. Where am I going wrong?
Reply 11
Is this OCR or OCR MEI, and which paper (year) was this? I haven't come across a matrices proof by induction before
Reply 12
Original post by Muffyn
Is this OCR or OCR MEI, and which paper (year) was this? I haven't come across a matrices proof by induction before


Ocr june 06
Reply 13
Original post by Super199
http://m.imgur.com/srgzWGU,J2u5gTV

9iii. Help me with the algebra pls haha. Where am I going wrong?


Your very first step. You pulled out a factor of 12(n+1)\frac{1}{2}(n+1), so you need to compensate by multiplying the relevant terms by 2.

i.e: 12(n+1)(2(n+1)223n)n1(n+1)\displaystyle \frac{1}{2}(n+1)\left(2(n+1)^2 - 2 - 3n\right) \underbrace{- n - 1}_{-(n+1)}

Looks like you need to put the maths away and rest for a while. :tongue:
(edited 7 years ago)
Reply 14
Original post by Zacken
Your very first step. You pulled out a factor of 12(n+1)\frac{1}{2}(n+1), so you need to compensate by multiplying the relevant terms by 2.

i.e: 12(n+1)(2(n+1)223n)n1(n+1)\displaystyle \frac{1}{2}(n+1)\left(2(n+1)^2 - 2 - 3n\right) \underbrace{- n - 1}_{-(n+1)}

Looks like you need to put the maths away and rest for a while. :tongue:


Yh got it lol. Im off to bed haha. Cheers
Reply 15
Original post by Super199
Yh got it lol. Im off to bed haha. Cheers


Nice, nice. Good night!

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