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    When n=k

    you sub in n for k

    so you get

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

    so when you use k+1 how would you do it?

    so far i have

    \displaystyle\sum_{r=1}^{k+1} r = \frac{k+1}{2} (k+1+1) ??
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    (Original post by huiop)
    When n=k

    you sub in n for k

    so you get

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

    so when you use k+1 how would you do it?

    so far i have

    \displaystyle\sum_{r=1}^(k+1) r = \frac{k+1}{2} (k+1+1) ??
    add the result for k and the (k+1) term together
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    (Original post by xyz9856)
    add the result for k and the (k+1) term together
    so in this case (k/2)(k+1)+(k+1)
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    Do what Sigma k equals plus k+1


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    (Original post by xyz9856)
    so in this case (k/2)(k+1)+(k+1)
    This


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    (Original post by xyz9856)
    add the result for k and the (k+1) term together
    so if i add the result for k then i have a 1 left right? but why add k+1 ???
    (Original post by xyz9856)
    so in this case (k/2)(k+1)+(k+1)
    ok so that's what you get when u do what you said....
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    THen add them together, and try and make it into what Sigma k+1 equals by factorising and stuff


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    (Original post by huiop)
    so if i add the result for k then i have a 1 left right? but why add k+1 ???


    ok so that's what you get when u do what you said....
    you add k+1 because if the summation formulae is to k+1 then its the sum to k plus the (k+1)th term, if that makes sense
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    (Original post by AdeptDz)
    THen add them together, and try and make it into what Sigma k+1 equals by factorising and stuff


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    ok ok i understand that i should add k+1 and when i do i can take out a factor of 0.5xk+1 and k+1+1
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    (Original post by xyz9856)
    you add k+1 because if the summation formulae is to k+1 then its the sum to k plus the (k+1)th term, if that makes sense
    and in this case the nth term is just r, so its k+1. If it was r^2+r+1 it would be sum to k + ((k+1)^2)+(k+1)+1
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    (Original post by huiop)
    When n=k

    you sub in n for k

    so you get

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

    so when you use k+1 how would you do it?

    so far i have

    \displaystyle\sum_{r=1}^{k+1} r = \frac{k+1}{2} (k+1+1) ??
    Prove that it now works for n=k+1

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

    QED.
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    (Original post by RDKGames)
    Prove that it now works for n=k+1

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

    QED.
    but shouldn't it be
    \displaystyle\sum_{r=1}^{k+1} r = \sum_{r=1}^k r + \sum_{r=1}^1 r ???
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    (Original post by huiop)
    but shouldn't it be
    \displaystyle\sum_{r=1}^{k+1} r = \sum_{r=1}^k r + \sum_{r=1}^1 r ???
    Of course not. That doesn't make sense. What you've shown says you are summing up integers from 1 up to k, and then adding an awkward 1 at the end.

    You need to add the (k+1)th term of the sequence because you are showing it works for (1+2+3+4+5+...+k)+(k+1). Since you are summing up integers from 1, the (k+1)th term will simply be (k+1)
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    With proof by induction you always have to use your assumption. In this case we are assuming that  \displaystyle \sum_{r=1}^n r = \frac{1}{2} n(n+1) . Using our assumption we have  \displaystyle \sum_{r=1}^{n+1} r = \frac{1}{2} n(n+1) + (n+1) .
    The way induction works is you assume that it works for some natural number  k and show that it implies that it works for the next natural number  k+1 . Then all you need to do is show that the result is indeed true for any value you pick and it completed the proof. Make sure you fully understand mathematical induction, I find that many people aren't convinced that it proves a result meaning that they do not properly understand it.
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    (Original post by huiop)
    but shouldn't it be
    \displaystyle\sum_{r=1}^{k+1} r = \sum_{r=1}^k r + \sum_{r=1}^1 r ???
    If you DO wish to express it as two different sums then:

    \displaystyle\sum_{r=1}^{k+1} r = \sum_{r=1}^k r + \sum_{r=k+1}^{k+1} r

    Notice the boundaries.
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    (Original post by RDKGames)
    Of course not. That doesn't make sense. What you've shown says you are summing up integers from 1 up to k, and then adding an awkward 1 at the end.

    You need to add the (k+1)th term of the sequence because you are showing it works for (1+2+3+4+5+...+k)+(k+1). Since you are summing up integers from 1, the (k+1)th term will simply be (k+1)
    ok thanks
    (Original post by B_9710)
    With proof by induction you always have to use your assumption. In this case we are assuming that  \displaystyle \sum_{r=1}^n r = \frac{1}{2} n(n+1) . Using our assumption we have  \displaystyle \sum_{r=1}^{n+1} r = \frac{1}{2} n(n+1) + (n+1) .
    The way induction works is you assume that it works for some natural number  k and show that it implies that it works for the next natural number  k+1 . Then all you need to do is show that the result is indeed true for any value you pick and it completed the proof. Make sure you fully understand mathematical induction, I find that many people aren't convinced that it proves a result meaning that they do not properly understand it.
    ah ok thanks i understand
    (Original post by RDKGames)
    If you DO wish to express it as two different sums then:

    \displaystyle\sum_{r=1}^{k+1} r = \sum_{r=1}^k r + \sum_{r=k+1}^{k+1} r

    Notice the boundaries.
    thanks
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    (Original post by B_9710)
    With proof by induction you always have to use your assumption. In this case we are assuming that  \displaystyle \sum_{r=1}^n r = \frac{1}{2} n(n+1) . Using our assumption we have  \displaystyle \sum_{r=1}^{n+1} r = \frac{1}{2} n(n+1) + (n+1) .
    The way induction works is you assume that it works for some natural number  k and show that it implies that it works for the next natural number  k+1 . Then all you need to do is show that the result is indeed true for any value you pick and it completed the proof. Make sure you fully understand mathematical induction, I find that many people aren't convinced that it proves a result meaning that they do not properly understand it.
    So is there anything i need to do to finalise the proof or does "therefore true for k+1 suffice"?
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    (Original post by huiop)
    So is there anything i need to do to finalise the proof or does "therefore true for k+1 suffice"?
    You usually need a final sentence saying:

    The result is true for n=1; therefore true for n=2,3,4,... by induction. QED.

    (QED is optional )
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    (Original post by huiop)
    So is there anything i need to do to finalise the proof or does "therefore true for k+1 suffice"?
    Just round it all off by saying something like result true for  n=k+1 if true for  n=k, k\in \mathbb{N} , result true for  n=1 \ \therefore true  \forall n, n\in \mathbb{N} .
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    (Original post by RDKGames)
    You usually need a final sentence saying:

    The result is true for n=1; therefore true for n=2,3,4,... by induction. QED.

    (QED is optional )
    ok thanks
 
 
 
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