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    I don't understand how to do proof by induction:
    the question im stuck on is;

    (the sum of) (3k+1) =1/2n(3n+5)

    PLEASE HELP SOMEONE
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    (Original post by Amyherdman)
    I don't understand how to do proof by induction:
    the question im stuck on is;

    (the sum of) (3k+1) =1/2n(3n+5)

    PLEASE HELP SOMEONE
    Have you tried it yet?
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    (Original post by edothero)
    Have you tried it yet?
    yes but i dont know how to do it
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    (Original post by Amyherdman)
    yes but i dont know how to do it
    Remember the 4 steps.
    Prove for n=1
    Assume true for n=k
    Let n=k+1 and prove
    Conclusion

    What step are you stuck on?
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    (Original post by edothero)
    Remember the 4 steps.
    Prove for n=1
    Assume true for n=k
    Let n=k+1 and prove
    Conclusion

    What step are you stuck on?
    mainly assume true for n=k
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    (Original post by Amyherdman)
    mainly assume true for n=k
    As far as I'm aware, for this part you just need to write a statement saying you assume its true for n=k? There's not much more you need to do for this step..
    Would be good if you could post your working out


    Zacken can take over here. I have a Physics exam to revise for
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    (Original post by Amyherdman)
    mainly assume true for n=k
    Yeah, all you need to do here is just write down Assume true for n=m, so that we have: \sum_{i=1}^k (3i+1) = \frac{1}{2}k(3k+5)

    That's it.

    (Original post by edothero)

    Zacken can take over here. I have a Physics exam to revise for
    Argh, trying to do S3 but I'll take over.
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    (Original post by edothero)
    As far as I'm aware, for this part you just need to write a statement saying you assume its true for n=k? There's not much more you need to do for this step..
    Would be good if you could post your working out

    Zacken can take over here. I have a Physics exam to revise for
    thank you, ive done that but thought i had to do more to help me prove n=k+1
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    (Original post by Amyherdman)
    mainly assume true for n=k
    Write the statement in k rather than n, not too difficult.
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    (Original post by Zacken)
    Yeah, all you need to do here is just write down Assume true for n=m, so that we have: \sum_{i=1}^k (3i+1) = \frac{1}{2}k(3k+5)

    That's it.



    Argh, trying to do S3 but I'll take over.
    what about n=k+1 then??
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    (Original post by zetamcfc)
    Write the statement in k rather than n, not too difficult.
    thats what i did but i thought there was more than that to do, how do you do n=k+1 then?
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    (Original post by Amyherdman)
    what about n=k+1 then??
    Then, for n=k+1, we want to prove that:

    \displaystyle \sum_{i=1}^{k+1} (3i+1) =^{??} \frac{1}{2}(k+1)(3(k+1) + 5) - we want to prove this equality somehow, so let's start from the sum:

    \displaystyle \sum_{i=1}^{k+1} (3i+1) = \sum_{i=1}^k (3i+1) + 3(k+1) - you've assumed something about the sum to k, maybe plug it in and do some factorising to prove what we want to prove?
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    (Original post by Amyherdman)
    thats what i did but i thought there was more than that to do, how do you do n=k+1 then?
    Zacken will help, got stuff to do.
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    Hint: The sum to k+1 = The sum to k + the (k+1)th term. Now substitute for the sum to k with the inductive hypothesis (the assumption that the statement is true for n=k).
 
 
 
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