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    I'm self-teaching FM AS and I'm fine with most of the things in FP1. But Mathematical Induction has sort of thrown me. Would someone please be able to just work this (supposedly) simple example and explain to me what you are doing?

    Thanks for you help :rolleyes:

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    Use the method of mathematical induction to prove that, for all positive integers n,

     1 \times 4 + 2 \times 5 + 3 \times 6 + ... + n(n + 3) = \frac{1}{3}n(n + 1)(n + 5)
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    wait a minute... am I doing something wrong, the base case N = 1 results in 4 for the lhs and 3 for the rhs, you might need to check you inputted it correct

    Edit - shouldn't it be 1/3 instead?
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    (Original post by Tobedotty)
    wait a minute... am I doing something wrong, the base case N = 1 results in 4 for the lhs and 3 for the rhs, you might need to check you inputted it correct

    Edit - shouldn't it be 1/3 instead?
    Yep, sorry. My mistake :cool:
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    The general idea of proof by induction in FP1 is:

    • State what you're trying to prove.
    • Show it's true for the appropriate base case.
    • Assume it's true for the general case n=k.
    • Show that it's true for the general "next term" n=k+1.
    • Since you have shown that it's true for the base case and the general next term, it must be true for all terms by induction.

    In your example:

    Spoiler:
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    Trying to prove that 1\times 4 + 2\times 5 + 3\times 6 + ... + n(n+3) = \frac{1}{3}n(n+1)(n+5). Call this (*).

    If n=1, LHS = 1*4 = 4 and RHS = 1*2*6/3 = 4. Hence true for n=1.

    Assume (*) is true for n=k. Then if n=(k+1):

    \mathrm{Sum} = \frac{1}{3}n(n+1)(n+5) + (n+1)(n+4) = (n+1)[\frac{1}{3}n(n+5) + n+4]
    \mathrm{Sum} = \frac{1}{3}(n+1)(n^2 + 8n + 12) = \frac{1}{3}(n+1)(n+2)(n+6) as required.

    Since (*) is true for n=1 and n=k+1, it's true for all integers n\geq 1.
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    (Original post by Smashing)
    I'm self-teaching FM AS and I'm fine with most of the things in FP1. But Mathematical Induction has sort of thrown me. Would someone please be able to just work this (supposedly) simple example and explain to me what you are doing?
    The principle is simple. Suppose you have some "sentence" P about positive integers. Then I'm going to write "P(n)" to mean "P is true for the number n". Then if you can prove that P(1), and that P(n) implies P(n+1), you've proven P is true for all positive integers. After all, we have P(1); and P(1) implies P(1+1), i.e. P(2); but then that in turn implies P(3), which implies P(4), which implies P(5), which...

    (Original post by Smashing)
    Use the method of mathematical induction to prove that, for all positive integers n,

     1 \times 4 + 2 \times 5 + 3 \times 6 + ... + n(n + 3) = \frac{1}{3}n(n + 1)(n + 5)
    Your sentence here is the above equality. It's true for n = 1 (you can verify this). Now suppose it's true for general 'n'; by adding (n+1)((n+1)+3) to both sides of this statement, show that this means it's also true for 'n+1'.
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    You could basically turn  1 \times 4 + 2 \times 5 + 3 \times 6 + ... + n(n + 3) = \frac{1}{3}n(n + 1)(n + 5)

    into :  \displaystyle\sum_{r=1}^ n n(n + 3) = \frac{1}{3}n(n + 1)(n + 5)

    Then follow the steps in the previous posts.
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    kinda off the subject, but is infinite descent like induction but from infinity down? IF not, the can induction be done in reverse? I.e n-1 rather than n+1?
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    (Original post by Tobedotty)
    kinda off the subject, but is infinite descent like induction but from infinity down?
    There's no such number as infinity. So no.

    (Original post by Tobedotty)
    IF not, the can induction be done in reverse? I.e n-1 rather than n+1?
    In some cases, yes. It wouldn't work here, though, because this statement isn't true for n = -1 (and doesn't even really make sense).
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    Given a starting point in the natural numbers you could certainly prove a proposition for all natural numbers below your starting number.

    (note: you don't actually start at infinity)
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    (Original post by generalebriety)
    There's no such number as infinity. So no.


    In some cases, yes. It wouldn't work here, though, because this statement isn't true for n = -1 (and doesn't even really make sense).
    I also forgot to ask what infinite descent is... do you know or should I just use google?
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    (Original post by Tobedotty)
    I also forgot to ask what infinite descent is... do you know or should I just use google?
    I do know what infinite descent is, but I don't see why you don't use google and then ask here if you get stuck. It'd make my job easier. :confused:

    The positive (or non-negative) integers are subject to the well-ordering principle; this means that, given any* set of positive integers, you can always find the smallest one. A proof by infinite descent goes like this:
    1. search for an integer solution to a problem
    2. show that, if a solution can be found, then a smaller one can always be found.
    The classic example is the proof of the irrationality of sqrt(2). Suppose you can write sqrt(2) = p/q (p, q positive integers); then you can always write sqrt(2) = m/n (m, n positive integers with m < p, n < q). Now keep doing this, and look at the sequence (p, m, ...) - this is a strictly decreasing sequence of positive integers which doesn't end, so violates the well-ordering principle.


    * non-empty, etc.
 
 
 
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