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Thread: induction

  1. MHB Craftsman

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    #1
    prove by induction for all positive integers n: 1+5+9+13+........+(4n-3)= n/2(4n-2)
    i tried this by trying to prove n/2(4n-2)+ (4(k+1)-3) = k+1/2(4(k+1)-2) but it did not work out for me.

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  3. Pessimist Singularitarian
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    #2
    Okay, so our induction hypothesis $P_n$ is:

    $ \displaystyle \sum_{k=1}^{n}(4k-3)=\frac{n}{2}(4n-2)=n(2n-1)$

    First we should verify that the base case $P_1$ is true:

    $ \displaystyle \sum_{k=1}^{1}(4n-3)=(1)(2(1)-1)$

    $ \displaystyle 4(1)-3=2-1$

    $ \displaystyle 1=1\quad\checkmark$

    Okay, the base case is true. So, for our induction step, let's write:

    $ \displaystyle \sum_{k=1}^{n}(4k-3)+4(n+1)-3=n(2n-1)+4(n+1)-3$

    Incorporating the added term into the sum on the left, we have:

    $ \displaystyle \sum_{k=1}^{n+1}(4k-3)=n(2n-1)+4(n+1)-3$

    So, what we need to do is demonstrate:

    $ \displaystyle n(2n-1)+4(n+1)-3=(n+1)(2(n+1)-1)$

    Can you proceed?

  4. MHB Craftsman

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    #3 Thread Author
    thanks

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    #4
    $ \displaystyle n(2n-1)+4(n+1)-3=2n^2+3n+1=(2n+1)(n+1)=(n+1)(2(n+1)-1)$

    Thus, we may state:

    $ \displaystyle \sum_{k=1}^{n+1}(4k-3)=(n+1)(2(n+1)-1)$

    This is $P_{n+1}$, which we have derived from $P_n$, thereby completing the proof by induction.

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