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    6ii.

    Which bit do I use?
    Un or Un+1 -Un?
    https://3ed6c47c6c1ead854571a830aaba...0FP1%20OCR.pdf

    Cheers.
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    (Original post by Super199)
    6ii.

    Which bit do I use?
    Un or Un+1 -Un?
    https://3ed6c47c6c1ead854571a830aaba...0FP1%20OCR.pdf

    Cheers.
    1. Show that u_1 is divisible by 2.
    2. Assume that u_n is divisible by 2.
    3. Show that u_{n+1} - u_{n} is divisible by 2 (this will require almost no work, just spot something from the first part).
    4. Conclude that u_{n+1} is divisible by 2.
    5. Write that induction thing.
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    (Original post by Zacken)
    1. Show that u_1 is divisible by 2.
    2. Assume that u_n is divisible by 2.
    3. Show that u_{n+1} - u_{n} is divisible by 2 (this will require almost no work, just spot something from the first part).
    4. Conclude that u_{n+1} is divisible by 2.
    5. Write that induction thing.
    I just did 2(n+2) so must be divisible by 2 haha. By the way how does 3 imply 4?
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    (Original post by Super199)
    I just did 2(n+2) so must be divisible by 2 haha. By the way how does 3 imply 4?
    Yep, that's correct.

    Because if u_{n+1} - u_n = m is divisible by 2 then add u_n to both sides to get u_{n+1} = m + u_n both terms are divisible by 2 (un is divisible by 2 by assumption) so the sum of two terms which are both divisible by 2 are also divisible by 2 hence u_{n+1} is divisible by 2.
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    (Original post by Zacken)
    Yep, that's correct.

    Because if u_{n+1} - u_n = m is divisible by 2 then add u_n to both sides to get u_{n+1} = m + u_n both terms are divisible by 2 (un is divisible by 2 by assumption) so the sum of two terms which are both divisible by 2 are also divisible by 2 hence u_{n+1} is divisible by 2.
    Cheers, mind helping with 4ii as well.

    I got 4i to be a circle centre (-1,1) radius root 2. I don't quite get the boundary thing with the second part, what am I looking for?
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    (Original post by Super199)
    Cheers, mind helping with 4ii as well.

    I got 4i to be a circle centre (-1,1) radius root 2. I don't quite get the boundary thing with the second part, what am I looking for?
    4(i) is asking the the locus of points that are exactly \sqrt(2) away from (1,-1) (surely it's (1, -1) and not (-1, 1))? That is the circle centre (1, -1) and radius \sqrt(2).

    4(ii) is asking for the locus of points that are \sqrt{2} or less away from (1, -1) but more than 1 away from (1, -1) which is precisely the shaded area inside of the circle between the two concentric circles of radius \sqrt(2) and radius 1.

    A diagram should help:

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    (Original post by Zacken)
    4(i) is asking the the locus of points that are exactly \sqrt(2) away from (1,-1) (surely it's (1, -1) and not (-1, 1))? That is the circle centre (1, -1) and radius \sqrt(2).

    4(ii) is asking for the locus of points that are \sqrt{2} or less away from (1, -1) but more than 1 away from (1, -1) which is precisely the shaded area inside of the circle between the two concentric circles of radius \sqrt(2) and radius 1.

    A diagram should help:

    Yh my bad with 4i.
    Thats one funky diagram I rate it. So you just want another circle (within the original one) with radius 1 from the centre and shade that bad boy?
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    (Original post by Super199)
    Yh my bad with 4i.
    Thats one funky diagram I rate it. So you just want another circle with radius 1 from the centre and shade that bad boy?
    You want to shade the area between the bad boy and the original circle. Basically the green lines on my diagram.
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    (Original post by Zacken)
    You want to shade the area between the bad boy and the original circle. Basically the green lines on my diagram.
    Yh got it cheers.
    8i, if you don't mind. I've forgotten how to deal with factorials lol
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    (Original post by Super199)
    Yh got it cheers.
    8i, if you don't mind. I've forgotten how to deal with factorials lol
    \displaystyle (r+2)! = (r+2)(r+1)\cdots (2)(1) = (r+2)(r+1)!

    So \displaystyle (r+2)! - (r+1)! = (r+2)(r+1)! - (r+1)! = \cdots - factorise that stuff and it'll fall out easily.
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    (Original post by Zacken)
    \displaystyle (r+2)! = (r+2)(r+1)\cdots (2)(1) = (r+2)(r+1)!

    So \displaystyle (r+2)! - (r+1)! = (r+2)(r+1)! - (r+1)! = \cdots - factorise that stuff and it'll fall out easily.
    I don't get what you have done sorry
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    (Original post by Super199)
    I don't get what you have done sorry
    (r+2)! is the same thing as (r+2) * (r+1) * (r) * (r-1) * (r-2) * ...

    (r+1)! is the same thing as (r+1)(r)(r-1)(r-2)(r-3)...

    The two bolded things are the same

    So I can rewrite (r+2)! as (r+2)(r+1)!.
 
 
 
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