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    The function, f(n), means the sum of the factors of n

     Show\ that\ f(2^n) = 1 + 2 + 2^2 + 2^3 + ... + 2^n

    I started by saying, x is a prime number who has factors 1 and x.
    x^n would have factors 1 and x^n and the number of factors would be equal to n + 1. (I don't know how to show this)

    Majorly messed up my Latex, was gona show what I've done so far, just adding it now...

    f(x^n) = 1 + x^n / x^n-1 + x^n / x^n-2 + x^n / x^n-3 + ... + x^n

    f (x^n) = 1 + x + x^2 + x^3 + ... + x^n

    then basically x = 2

    But I know I haven't SHOWN anything but I can't think of anything else
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    Then it says find an expression for a prime number p for f(p^n)

    And show that is equals  (p^{n+1} - 1) / p - 1
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    So yeah thanks a lot for anyone who can help me out here plus rep obvs.
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    I'm not sure how you've done that to be honest but could it not be proved fairly simply by induction?
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    (Original post by corney91)
    I'm not sure how you've done that to be honest but could it not be proved fairly simply by induction?
    I assumed there would be an easier way than induction because it's from a GCSE level text book.
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    (Original post by SahilB91)
    x^n would have factors 1 and x^n
    Are you still assuming x is prime here? Because then x is a root of x^n... I'm not sure what you're trying to say... :confused:

    (Original post by SahilB91)
    I assumed there would be an easier way than induction because it's from a GCSE level text book.
    Yeah, induction is definitely not in the GCSEs but this looks way harder than anything I ever did at GCSE :eek:
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    The question you need to answer: What are the factors of 2^n?

    It's easy to "see" what they are, and I'm not sure you really need to justify it - particularly at GCSE. But if you do want to prove it: Suppose k divides 2^n. Let p be a prime dividing k. Can p be anything other than 2? So, we can write k in the form 2^m for some m. And then m must be <= n.

    For the 2nd problem, at the end:

    Spoiler:
    Show
    You will want to show that 1+p^2+...+p^n = \frac{p^{n+1}-1}{p-1}. Easiest way to do this is multiply both sides by (p-1) and notice that an awful lot of stuff on the LHS cancels.
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    (Original post by corney91)
    Are you still assuming x is prime here? Because then x is a root of x^n... I'm not sure what you're trying to say... :confused:


    Yeah, induction is definitely not in the GCSEs but this looks way harder than anything I ever did at GCSE :eek:
    Yeah it is a bit abstract. Well I don't think x is always a root of n whereas I think x^n always is. Because with this function say n = 1 then the same factor would be summed twice, but I may be wrong. So I think as long as n > 2 then x would be.
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    (Original post by DFranklin)
    The question you need to answer: What are the factors of 2^n?

    It's easy to "see" what they are, and I'm not sure you really need to justify it - particularly at GCSE. But if you do want to prove it: Suppose k divides 2^n. Let p be a prime dividing k. Can p be anything other than 2? So, we can write k in the form 2^m for some m. And then m must be <= n.

    For the 2nd problem, at the end:

    Spoiler:
    Show
    You will want to show that 1+p^2+...+p^n = \frac{p^{n+1}-1}{p-1}. Easiest way to do this is multiply both sides by (p-1) and notice that an awful lot of stuff on the LHS cancels.
    I really like the first bit you wrote. I am not to sure about the spoiler contents because:

    multiplying both sides by (p-1) seems to always leave something like...

    p = p^n :confused:

    1 + p^n = (p^n+1 - 1) / p -1

    p + p^n+1 - 1 - p^n = p^n+1 - 1

    p - p^n = 0

    p = p^n
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    (Original post by SahilB91)
    I really like the first bit you wrote. I am not to sure about the spoiler contents because:

    multiplying both sides by (p-1) seems to always leave something like...

    p = p^n :confused:

    1 + p^n = (p^n+1 - 1) / p -1

    p + p^n+1 - 1 - p^n = p^n+1 - 1

    p - p^n = 0

    p = p^n
    I don't see where any of these lines have come from. I think you are making some algebra errors. Write it down for some simple case - n = 2, say.
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    (Original post by DFranklin)
    I don't see where any of these lines have come from. I think you are making some algebra errors. Write it down for some simple case - n = 2, say.
    Sorry, if you look at your earlier post, although multiplying both sides by (p-1) should work it doesn't in practice.

    As you always end up with p^n = p^a

    where a is dependant on the highest power of p in the equation.

    i.e. if there a p^3 then a = 3, if there is just p then a = 1.

    I am really stuck here because it works when there is a n = any number, but I can't show with n. :mad:
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    It does work. I suggest you post your working.
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    (Original post by DFranklin)
    It does work. I suggest you post your working.
    Right ok (thanks btw).

    This is where I am going wrong...

    When it says show that  1 + p + p^2 + ... + p^n = \frac{p^{n+1} - 1}{p-1}

    I am not sure about the LHS and what to leave out when I multiply by (p-1).

    I mean do I multiply (p-1) against the whole LHS how it appears now, or do I leave out the p^2 term and only use 1 + p + p^n? :confused:
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    The whole LHS. (I don't understand how you could think you could leave anything out without ending up with the wrong answer).
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    (Original post by SahilB91)
    Sorry, if you look at your earlier post, although multiplying both sides by (p-1) should work it doesn't in practice.

    As you always end up with p^n = p^a

    where a is dependant on the highest power of p in the equation.

    i.e. if there a p^3 then a = 3, if there is just p then a = 1.

    I am really stuck here because it works when there is a n = any number, but I can't show with n. :mad:

    It does work. You can easily prove it by induction using the formula that DFranklin gave you (S= [(p^n+1)-1]/(p-1) for S=1+p+p²+...+p^n )

    or you can just multiply by (p-1).

    Another proof, just so that you are convinced, is to do it this way:

    For n a fixed positive integer, you have:

    Write down your expression S(n) = 1 + p +...+p^n
    multiply by p, and you get pS(n) = p + p² +...+p^n + p^(n+1)

    Substract S(n)-pS(n)
    If you write S(n) and pS(n) one under the other, you will easily see that in the substraction, every term cancels out, EXCEPT 1 and p^n+1

    Therefore, S(n)-pS(n) = 1 - p^(n+1)
    Therfore, S(n) = (1-p^n+1)/(1-p)

    Multiply again by -1, and you get the formula that DFranklin gave you
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    Sigh. A perfect example of someone jumping in when I was waiting for the OP to try to do it himself.
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    (Original post by DFranklin)
    The whole LHS. (I don't understand how you could think you could leave anything out without ending up with the wrong answer).
    This is why:

     1 + p + p^2 + ... + p^n = \frac{p^{n+1} - 1}{p-1}

     (1 + p + p^2 + p^n)(p-1) = p^{n+1} - 1

     p + p^2 + p^3 + p^{n+1} - 1 - p - p^2 - p^n = p^{n+1} - 1

    p^3 - p^n = 0

    p^3 = p^n

    Is my algebra wrong? :confused:
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    In "1 + p + p^2 + ... + p^n", the "..." means 'and so on' - that is, that you continue the pattern until you get to the end term.

    In other words: "1 + p + p^2 + ... + p^5" = "1 + p + p^2 + p^3 + p^4 + p^5".
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    (Original post by paronomase)
    It does work. You can easily prove it by induction using the formula that DFranklin gave you (S= [(p^n+1)-1]/(p-1) for S=1+p+p²+...+p^n )

    or you can just multiply by (p-1).

    Another proof, just so that you are convinced, is to do it this way:

    For n a fixed positive integer, you have:

    Write down your expression S(n) = 1 + p +...+p^n
    multiply by p, and you get pS(n) = p + p² +...+p^n + p^(n+1)

    Substract S(n)-pS(n)
    If you write S(n) and pS(n) one under the other, you will easily see that in the substraction, every term cancels out, EXCEPT 1 and p^n+1

    Therefore, S(n)-pS(n) = 1 - p^(n+1)
    Therfore, S(n) = (1-p^n+1)/(1-p)

    Multiply again by -1, and you get the formula that DFranklin gave you
    I feel so stupid because I've done further maths and I didn't even consider induction because this is from a GCSE textbook. :mad:
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    (Original post by DFranklin)
    In "1 + p + p^2 + ... + p^n", the "..." means 'and so on' - that is, that you continue the pattern until you get to the end term.

    In other words: "1 + p + p^2 + ... + p^5" = "1 + p + p^2 + p^3 + p^4 + p^5".
    Yeah I shouldn't have wasted so much time and just used a multiplicative factor :facepalm:
 
 
 
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