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    I've just started learning mathematical induction and I'm at a loss, I know what we're trying to do and I *sort of* know how to go about it but I don't know what the end result should look like. From what I've read, all end results have (k+1) in them.

    What I do know:

    - How to add the k+1 value on the RHS

    What I don't know:

    - How the end result should look like. Does just having the value (k+1) in it make it correct?
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    Say one were trying to prove that,

    \displaystyle \sum_{r=1}^{n} r = 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}

    one would assume

    \displaystyle \sum_{r=1}^{k} r = 1 + 2 + 3 + \dots + k = \frac{k(k+1)}{2}

    and the result of the inductive step would look like

    \displaystyle \sum_{r=1}^{k+1} r = 1 + 2 + 3 + \dots + k + (k+1) = \frac{(k+1)((k+1)+1)}{2}

    (i.e. replace all instances of k with k+1)

    The whole idea is that if one knows a statement is true for a particular integer, such as 1, and one can prove that if true for one integer, it's also true for the next integer, then since we know it is true for 1, it follows that it is also true for 2, and for 3, and for 4, and so on...

    So in other words, we're always interested in getting from k to k+1 because that allows us set up a kind of cascade if you will, through all the integers.
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    OK, tell me, is this correct?

    \sum_{r=1}^{k}2r = 2+4+6+...+(2k) = k(k+1)

\\\\

\sum_{r=1}^{k+1}2r = 2+4+6+...+(2k)+2(k+1) = k(k+1)+2(k+1)

\\\\

= k^2+k+2k+2\\\\

=k^2+3k+2\\\\

=(k+1)(k+2)

    Sorry for the incorrect use of sigma, still getting used to latex.
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    (Original post by mcp2)
    OK, tell me, is this correct?

    \sum_{r=1}^{k}r = 2+4+6+...+(2k) = k(k+1)

\\\\

\sum_{r=1}^{k+1}r = 2+4+6+...+(2k)+2(k+1) = k(k+1)+2(k+1)

\\\\

= k^2+k+2k+2\\\\

=k^2+3k+k\\\\

=(k+1)(k+2)

    Sorry for the incorrect use of sigma, still getting used to latex.
    I have corrected a couple of typing errors.

    \sum_{r=1}^{k}2r = 2+4+6+...+(2k) = k(k+1)

\\\\

\sum_{r=1}^{k+1}2r = 2+4+6+...+(2k)+2(k+1) = k(k+1)+2(k+1)

\\\\

= k^2+k+2k+2\\\\

=k^2+3k+2\\\\

=(k+1)(k+2)
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    Thank you, edited. So is it correct?
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    Yep, looks correct to me.
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    To make the sums look better, you may want to use \displaystyle before it. Like this:

    [latex]\displaystyle\sum_{r=1}^k r[/latex] gives \displaystyle\sum_{r=1}^k r
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    I would lay it like this ( not that it really makes any difference, just habit)




    \displaystyle\sum_{r=1}^{k}2r = k(k+1)

\\\\\

\sum_{r=1}^{k}2r+2(k+1) = k(k+1)+2(k+1)

\\\\\

\sum_{r=1}^{k+1}2r= k(k+1)+2(k+1)

\\\\\

= k^2+k+2k+2\\\\

=k^2+3k+2\\\\

=(k+1)(k+2)\\\\

=(k+1)((k+1)+1)
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    OP, am I right in saying that you were expressing an appeal for insight on induction in general, and not just a specific example?

    I forget who told me this way of thinking it, but this is a fairly useful structure for basing inductive arguments around.

    Let's say we want to climb an infinite ladder. We want to show that we can climb as far up as we like (so we want to show that a claim holds for any k). We can do this by showing three things:

    1) We can get on to some rung of the ladder. This is the 'starting point' or 'basis' of your proof. It doesn't matter where you start on the ladder (unless you are asked to show something works from a specific stage!).

    Now we want to show that if we're standing on one step, we can climb to the next step up.

    2) We assume that we can stand on the k'th rung of the ladder.

    3) Using the assumption that you can stand on the k'th step, you want to show that you can climb to the next step.

    Now, if you know that you can find a rung that you can stand on, and that if you are on a rung, you can go to the next one up, it's obvious that you can go arbitrarily high on the ladder. In terms of your proof, you would structure it like:

    Base case: 'Such-and-such' holds for this particular value of k. Look, here is a computation: (...do a computation )

    Suppose that 'Such-and-such' holds for a number 'k'.
    (Then you use this assumption to show things work for k+1)
    So 'Such-and-such' holds for all k bigger than the k we found in the base case.

    I apologise if that was useless and a waste of space and time :o:
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    There is also a principle of strong induction, where you assume the truth of a proposition for all k=<n (less than or equal to n), and then use this to prove the truth of k+1.
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    To be honest you have to play around with proving other examples to get use to it. Certainly, it didn't click for me intill I did lots of example of induction.

    Prove
    (1+x)^n \geq 1+nx
    for all non negative integers n and real numbers x>-1.

    If you can do this without help then you understand induction.
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    No worries, it's good to get a different explanation apart from the domino one you hear all the time

    (Original post by Simplicity)
    Prove
    (1+x)^n \geq 1+nx
    for all non negative integers n and real numbers x>-1.
    Basis case for n = 0
    (1+x)^0 \geq 1+0x\\1 = 1

    For n = k+1
    (1+x)^{k+1} \geq 1+(k+1)x\\(1+x)^k(1+x)^1 \geq 1+kx+x\\(1+x)^k(1+x) \geq 1+kx+x

    I get stuck on the (1+x)^k bit, I've seen in other proofs that (1+x)^k = k+kx but I don't understand how that could be?

    I think I've gone very wrong (1+x)^k isn't the same as 1^k+x^k
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    weak:
    1. show (0)
    2. show that if (n) then (n+1)

    strong:
    1. show (0)
    2. show that if (0),(1),...(n) then (n+1)

    It's probably one of the hardest things you do in maths when you consider the stage at which you're expected to learn it. From experience, practice does not exactly make perfect (as some induction proofs are very hard) but it makes you pretty damn good at answering exam questions.
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    (Original post by mcp2)
    I get stuck on the bit, I've seen in other proofs that x^k = kx but I don't understand how that could be?

    I think I've gone very wrong (1+x)^k isn't the same as 1^k+x^k
    It looks like you don't understand induction

    Start with n=k and try to manipulate it to get n=k+1,
    • Thread Starter
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    Thinking about it understand how (1+x)^k becomes 1^k+x^k but isn't there something missing?

    I tried to do it for k, but as I said I'm not sure where to go from (1+x)^k.
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    from (1+x)^k >= 1+kx, multiply both sides by (1+x).
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    (Original post by DFranklin)
    from (1+x)^k >= 1+kx
    He is never going to learn induction with such a big hint. You might aswell do the working out for him to.
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    Wiki Support Team
    (Original post by mcp2)
    For n = k+1
    (1+x)^{k+1} \geq 1+(k+1)x
    No, this is what you're trying to prove. You don't know this yet. All you know is that (1+x)^n \geq 1+nx for n \leq k. So the sensible thing to do is to put n = k and then multiply both sides by (1+x).
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    (Original post by Simplicity)
    He is never going to learn induction with such a big hint. You might aswell do the working out for him to.
    The principles of induction have nothing to do with the specific mechanics of completing the induction step, so your post is unsupportable.
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    I don't know, wouldn't it be better for the OP to do induction by himself. Obviously, he knows the principle of it. But, then you have to build up intuition by working it out yourself.
 
 
 
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