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    I'm a bit confused with the following question:

    Porve by mathematical induction that sigma r = 1/2n(n+1)

    I understand all the working out up until:

    When n=k+1

    sigma r = 1/2k(k+1) + (k+1)
    =1/2(k+1)(k+2)
    =1/2(k+1)(k+1)

    ^ I'm confused on how ot get from the first line to the second line?

    Thanks in advance.
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    Your first line is correct, and the final answer. It is what \frac{1}{2} n(n+1) equals when n=k+1.
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    (Original post by Hopple)
    Your first line is correct, and the final answer. It is what \frac{1}{2} n(n+1) equals when n=k+1.

    I know the first line is correct but I don't know how to get from that first line to the second line?
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    You don't want the second line, throw it away

    What you've done in getting the first line is show that if \sum_1^n r=\frac{1}{2} n(n+1) is true for n=k, then it must also be true for n=k+1, which is your inductive step.
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    (Original post by Hopple)
    You don't want the second line, throw it away

    What you've done in getting the first line is show that if \sum_1^n r=\frac{1}{2} n(n+1) is true for n=k, then it must also be true for n=k+1, which is your inductive step.
    Yes but how did it get from 1/2k (k+1) + (k-1)

    to 1/2 (k+1)(k+2)
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    (Original post by Kosovarja94)
    Yes but how did it get from 1/2k (k+1) + (k-1)

    to 1/2 (k+1)(k+2)
    do you mean 1/2k (k+1) + (k+1) ?
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    Take (k+1) out as a factor:

    (k+1) ( 1/2k + 1)
    = (k+1)*(1/2)(k+2)
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    (Original post by kfkle)
    do you mean 1/2k (k+1) + (k+1) ?
    Sorry, yes I mean 1/2k (k+1) (k+1)
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    So you're having difficulty with thse two lines?

    sigma r = 1/2k(k+1) + (k+1)
    =1/2(k+1)(k+2)

    Just stick both terms in the same fraction (over 2) and then factorise the quadratic.
 
 
 
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