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    I'm not really sure where to begin with this. I assumed true for n=3 into both sequences of a(n) which I got 3/2 for each for so therefore true. The part which is confusing is finding n=k+1 since I'm not sure which sequence of a(k) I need to use. Is it possible to make the two sequences equal to each other solve that way?
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    (Original post by As_Dust_Dances_)
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    Your hypothesis should be for k and k+1. Then using the recurrence relation, show that it is true for k+2.
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    (Original post by Lord of the Flies)
    Your hypothesis should be for k and k+1. Then using the recurrence relation, show that it is true for k+2.
    I seem to get

    2 [1 + (-0.5)^k+1 + (-0.5)^k]

    but I'm not sure how this simplifies to

    2 + 4(-0.5)^k+2
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    (Original post by Lord of the Flies)
    Your hypothesis should be for k and k+1. Then using the recurrence relation, show that it is true for k+2.
    I think I've done it, is there a part where you let (-0.5)^2 = (0.5)^2 by any chance?
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    (Original post by As_Dust_Dances_)
    I think I've done it, is there a part where you let (-0.5)^2 = (0.5)^2 by any chance?
    First, you want to verify the equation actually works. So verify that

    a_2 = 2 + 4(-\frac{1}{2})^2 = 2 + 1 = 3

    So it works for the base case

    Now, you want to assume that it's true for some n=k. So essentially you're assuming the following:

    a_k = 2 + 4 \left(-\frac{1}{2}\right)^k

    and we also know from the recurrence relationship:

    a_k = \frac{1}{2} \left(a_{k-1}+a_{k-2} \right)

    Now, in induction (this is technically called strong induction although the naming does seem to imply this is 'strong' - it's actually no different from normal induction) you assume it's true for n=k, but you can also assume it's true for n=p where p<k. So essentially, we can use the inductive hypothesis for n=k, n=k-1, n=k-2 , ... , n=2

    Ok so, look at the case where n=k+1 and we want to get this case into the format of the equation

    a_n = 2 + 4 \left(-\frac{1}{2}\right)^n

    So what we're actually looking to show, is that:

    a_{k+1} = 2 + 4 \left(-\frac{1}{2}\right)^{k+1}

    If you struggle with induction, it is normally a very good idea to write down what it is you're trying to end up with (also it can be very useful to note how else it could be written) so, for example, the above equation is equivalent to:

    a_{k+1} = 2 + 4\left(-\frac{1}{2}\right) \left(-\frac{1}{2}\right)^{k} = 2 - 2 \left(-\frac{1}{2}\right)^{k}

    Anyway, we know from the recurrence relationship that:

    a_{k+1} = \frac{1}{2}\left(a_k + a_{k-1}\right) \ \ \ (*)

    Now, we already have assumed that

    a_k = 2 + 4 \left(-\frac{1}{2}\right)^k

    and using strong induction:

    a_{k-1} = 2 + 4 \left(-\frac{1}{2}\right)^{k-1}

    To finish this you need to substitute both of these into (*) and also to note that

    \left(-\frac{1}{2}\right)^{k-1}\ = 4\left(-\frac{1}{2}\right)^{k+1}


    EDIT: Apologies if I've explained the obvious to you, I'm just trying to avoid my own work!
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    (Original post by Noble.)
    First, you want to verify the equation actually works. So verify that

    a_2 = 2 + 4(-\frac{1}{2})^2 = 2 + 1 = 3

    So it works for the base case

    Now, you want to assume that it's true for some n=k. So essentially you're assuming the following:

    a_k = 2 + 4 \left(-\frac{1}{2}\right)^k

    and we also know from the recurrence relationship:

    a_k = \frac{1}{2} \left(a_{k-1}+a_{k-2} \right)

    Now, in induction (this is technically called strong induction although the naming does seem to imply this is 'strong' - it's actually no different from normal induction) you assume it's true for n=k, but you can also assume it's true for n=p where p<k. So essentially, we can use the inductive hypothesis for n=k, n=k-1, n=k-2 , ... , n=2

    Ok so, look at the case where n=k+1 and we want to get this case into the format of the equation

    a_n = 2 + 4 \left(-\frac{1}{2}\right)^n

    So what we're actually looking to show, is that:

    a_{k+1} = 2 + 4 \left(-\frac{1}{2}\right)^{k+1}

    If you struggle with induction, it is normally a very good idea to write down what it is you're trying to end up with (also it can be very useful to note how else it could be written) so, for example, the above equation is equivalent to:

    a_{k+1} = 2 + 4\left(-\frac{1}{2}\right) \left(-\frac{1}{2}\right)^{k} = 2 - 2 \left(-\frac{1}{2}\right)^{k}

    Anyway, we know from the recurrence relationship that:

    a_{k+1} = \frac{1}{2}\left(a_k + a_{k-1}\right) \ \ \ (*)

    Now, we already have assumed that

    a_k = 2 + 4 \left(-\frac{1}{2}\right)^k

    and using strong induction:

    a_{k-1} = 2 + 4 \left(-\frac{1}{2}\right)^{k-1}

    To finish this you need to substitute both of these into (*) and also to note that

    \left(-\frac{1}{2}\right)^{k-1}\ = 4\left(-\frac{1}{2}\right)^{k+1}


    EDIT: Apologies if I've explained the obvious to you, I'm just trying to avoid my own work!
    Thanks a lot for your help! I wrote down your method too, but instead I assumed true for n=k+2 just to get the recurrence relationship in terms of k+1 and k:-

    a_{k+2} = \frac{1}{2}\left(a_{k+1} + a_k\right) \ \ \ (*)

    Although your working is probably more simpler :L
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    (Original post by As_Dust_Dances_)
    Thanks a lot for your help! I wrote down your method too, but instead I assumed true for n=k+2 just to get the recurrence relationship in terms of k+1 and k:-

    a_{k+2} = \frac{1}{2}\left(a_{k+1} + a_k\right) \ \ \ (*)

    Although your working is probably more simpler :L
    Yeah, sometimes it can be easier to use n=k+2 with recurrence relations - either way it doesn't really matter

    Glad I could help.
 
 
 
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