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Math Help - Mathematical Induction problem

  1. #1
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    Mathematical Induction problem

    problem #14 on here:



    This is what ive done so far:
    1.) prove that the 2 equations are equal by testing P(n) at 0. They are, so i proceed to the inductive step and rewrite the equation in terms of k.

    2.) I then try nd show for all integers > or equal to k if P9K) is true then P(k+1) is also true.

    I do this by equating \left( k+1 \right)2^{\left( k+1 \right)+2}+2

    with \sum_{i=0}^{k+1}{k2^{k}}+k^{k+1}

    then attempt to turn it into what got for P(k+1), but im pretty sure i messed up making the last term explicit.
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  2. #2
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    Quote Originally Posted by Evan.Kimia View Post
    problem #14 on here:



    This is what ive done so far:
    1.) prove that the 2 equations are equal by testing P(n) at 0. They are, so i proceed to the inductive step and rewrite the equation in terms of k.

    2.) I then try nd show for all integers > or equal to k if P9K) is true then P(k+1) is also true.

    I do this by equating \left( k+1 \right)2^{\left( k+1 \right)+2}+2

    with \sum_{i=0}^{k+1}{k2^{k}}+k^{k+1}

    then attempt to turn it into what got for P(k+1), but im pretty sure i messed up making the last term explicit.
    So just to be clear after checking the bases case
    we assume the induction hypothesis for n=k this gives

    \displaystyle \sum_{i=1}^{k+1}i2^i=k2^{k+2}+2

    Now we need to use this to show k+1 starting with the left hand side we get

    \displaystyle \sum_{i=1}^{(k+1)+1}i2^i=\displaystyle \sum_{i=1}^{k+2}i2^i

    Now lets pull the last term out of the sum to get

    \displaystyle (k+2)2^{k+2} +\sum_{i=1}^{k+1}i2^i

    Now the sum is the induction hypothesis so we replace it with

    \displaystyle (k+2)2^{k+2} +k2^{k+2}+2=[(k+2)+k]2^{k+2}+2=
    \displaystyle [2k+2]2^{k+2}+2=2(k+1)2^{k+2}+2=(k+1)2^{k+3}+2 =(k+1)2^{(k+1)+2}+2
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  3. #3
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    perfect explanation, thanks!
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  4. #4
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    Quote Originally Posted by Evan.Kimia View Post
    problem #14 on here:


    This is what ive done so far:
    1.) prove that the 2 equations are equal by testing P(n) at 0. They are, so i proceed to the inductive step and rewrite the equation in terms of k.

    2.) I then try nd show for all integers > or equal to k if P9K) is true then P(k+1) is also true.

    I do this by equating \left( k+1 \right)2^{\left( k+1 \right)+2}+2

    with \sum_{i=0}^{k+1}{k2^{k}}+k^{k+1} this part is incorrect

    then attempt to turn it into what got for P(k+1), but im pretty sure i messed up making the last term explicit.
    P(k) is

    \displaystyle\sum_{i=0}^{k+1}i2^i=k2^{k+2}+2


    Then P(k+1) is

    \displaystyle\sum_{i=0}^{k+2}i2^i=(k+1)2^{k+3}+2

    To prove that the Domino-effect exists
    true for n=0 implies true for n=1
    true for n=1 implies true for n=2
    true for n=2 implies true for n=3
    true for n=3 implies true for n=4
    to infinity.

    Hence we try to show that P(k) being true forces P(k+1) to be true..

    \displaystyle\sum_{i=0}^{k+2}i2^i=\sum_{i=0}^{k+1}  i2^i+(k+2)2^{k+2}

    and "if" P(k) is true, this will be

    k2^{k+2}+2+(k+2)2^{k+2}

    =k2^{k+2}+2+k2^{k+2}+2^{k+3}

    =2k2^{k+2}+2^{k+3}+2=k2^{k+3}+2^{k+3}+2

    =(k+1)2^{k+3}+2
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