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    https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf

    9ii


    f(k+1)-f(k)= 4^k+1 + 6(k+1) + 8 - (4^k +6k +8)

    I don't know where to go from here, anyone help please
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    Try expanding and simplifying to the best of your ability, then try and look for common terms to factorise parts of it.
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    (Original post by lukejoshjames)
    Try expanding and simplifying to the best of your ability, then try and look for common terms to factorise parts of it.
    I get 3(4^k) + 6 I suspect that I have gone wrong? :/
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    (Original post by Damien_Dalgaard)
    I get 3(4^k) + 6 I suspect that I have gone wrong? :/
    Yeah, you need to leave some thing not simplified I think, as you need to find f(k) within the expression.

    At the moment I have f(k+1)-f(k) = f(k)+4(4^k)-6k-7, but I think that may be wrong too.
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    (Original post by Damien_Dalgaard)
    I get 3(4^k) + 6 I suspect that I have gone wrong? :/
    thats right man. you can finish it easily from there. just use modular arithmetic: consider the remainder when you divide 4^k by 18, and observe that it cycles through a fixed set of integers. Then prove it cycles through thsi fixed set of integers, and show that they are always congruent to 12mod18 when multiplied by 3, and then you are done
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    (Original post by Damien_Dalgaard)
    I get 3(4^k) + 6 I suspect that I have gone wrong? :/
    Give me like 5 minutes to screw about with Latex?
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    (Original post by Jai Sandhu)
    Give me like 5 minutes to screw about with Latex?

    You legend
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    (Original post by CancerousProblem)
    thats right man. you can finish it easily from there. just use modular arithmetic: consider the remainder when you divide 4^k by 18, and observe that it cycles through a fixed set of integers. Then prove it cycles through thsi fixed set of integers, and show that they are always congruent to 12mod18 when multiplied by 3, and then you are done
    That is a bit much for FP1 no? This is not STEP
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    (Original post by Jai Sandhu)
    That is a bit much for FP1 no? This is not STEP
    i haven't done fp1, but I tried the problem anyway and that's exactly what I did. I don't see an easier way out if you have to use induction.

    Alternately you can just use induction again on 3(4^k)+6 to show it's always divisible by 18,
    3(4^(k+1)+6
    =3(4^k)+6 + 3(3)(4^k)
    =3(4^k)+6 + 9(2^k)(2^k) which is obviously divisible by 18
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    (Original post by CancerousProblem)
    i haven't done fp1, but I tried the problem anyway and that's exactly what I did. I don't see an easier way out if you have to use induction
    There is, writing it up atm.
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    (Original post by Jai Sandhu)
    There is, writing it up atm.
    Induction twice is not easier than modular arithmetic lol
    modular arithmetic is really straightforward to use here, can't see a reason not to
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    (Original post by CancerousProblem)
    Induction twice is not easier than modular arithmetic lol
    modular arithmetic is really straightforward to use here, can't see a reason not to
    He wont get the marks :/

    Btw sorry for delay, latex is killing me T_T
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    (Original post by Jai Sandhu)
    He wont get the marks :/

    Btw sorry for delay, latex is killing me T_T
    seriously? modular arithmetic banned in FP1? Are you kidding lol...

    I hope step doesn't ban AM-GM since their questions use it so much, it would be annoying to have to prove AM-GM everytime i want to use it

    edit: yes the question said to use induction, but you can induct once and once you get the 3(4^k)+6 component you can use modular arithmetic to sort it out. That would still be induction, right?
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    (Original post by Damien_Dalgaard)
    https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf

    9ii


    f(k+1)-f(k)= 4^k+1 + 6(k+1) + 8 - (4^k +6k +8)

    I don't know where to go from here, anyone help please
    4n
    + 6n + 8

    



f(k) = 4^k + 6k + 8



f(k+1) + f(k) = 4^{k+1} + 6(k+1) + 8 + 4^(k) + 6k + 8



f(k+1) + f(k) = 4^k +4 \times 4^k + 6(k+1) +6k + 8 + 8



f(k+1) + f(k) = 5 \times 4^k + 12k + 24



f(k+1) = 5 \times 4^k + 12k + 22 - f(k) 



f(k+1) = 5 \times 4^k + 12k + 22 - (4^k + 6k + 8)



f(k+1) = 4 \times 4^k + 6k + 14



f(k+1) = 4 \times (4^k + 6k + 8) - 18k - 18



f(k+1) = 4 \times f(k) - 18(k - 1)

    I think you can do the rest?
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    Didn't you already get A in further maths?
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    (Original post by Jai Sandhu)
    4n
    + 6n + 8

    removed latex quote
    I think you can do the rest?
    meh honestly tat looks more complicated to me but that's just my opinion
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    OK I fixed the latex!!!! Goddamnit.
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    (Original post by CancerousProblem)
    meh honestly tat looks more complicated to me but that's just my opinion
    Can you requote that with the edited version, I published a latex form which is a bit broken... thanks. Yeah it is more complicated but hey, its what the examiners want.
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    (Original post by CancerousProblem)
    meh honestly tat looks more complicated to me but that's just my opinion
    It is more complicated, as you have to find exactly what THEY want, and not the way you want to do it. Sometimes they have alternative methods but they're mostly as complicated.
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    (Original post by CancerousProblem)
    seriously? modular arithmetic banned in FP1? Are you kidding lol...

    I hope step doesn't ban AM-GM since their questions use it so much, it would be annoying to have to prove AM-GM everytime i want to use it

    edit: yes the question said to use induction, but you can induct once and once you get the 3(4^k)+6 component you can use modular arithmetic to sort it out. That would still be induction, right?
    I think it would be sketchy, STEP allows any innovative idea so you will be set for that
 
 
 

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