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Math Help - Help again on mathematical induction. Also checking an answer as well.

  1. #1
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    Help again on mathematical induction. Also checking an answer as well.

    First, here's the answer I want to check.

    \frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}  {(2n-1)(2n+1)}=\frac{n}{2n+1}
    s_1=\frac{1}{3} (check)
    \frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}  {(2n-1)(2n+1)}+\frac{1}{(2k+1)(2k+3)}
    =P_k+\frac{1}{(2k+1)(2k+3)}
    =\frac{k}{2k+1}+\frac{1}{(2k+1)(2k+3)}
    =\frac{k+1}{2k+1}

    ~~~~~~~~~~~~~~~~~~~

    Second, on the problem:

    2+7+...+(5n-3)=\frac{n}{2}(5n-1)

    I get to:

    P_k+(5k+2)
    \frac{k}{2}(5k+1)+(5k+2)

    And once again, I get stuck trying to finish the problem off. I know how to get to this point, I just fail to see how to finish it off

    Any and all help (and verification for the first problem) will be greatly appreciated.
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  2. #2
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    Quote Originally Posted by mathgeek777 View Post
    First, here's the answer I want to check.

    \frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}  {(2n-1)(2n+1)}=\frac{n}{2n+1}
    s_1=\frac{1}{3} (check)
    \frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}  {(2n-1)(2n+1)}+\frac{1}{(2k+1)(2k+3)}
    =P_k+\frac{1}{(2k+1)(2k+3)}
    =\frac{k}{2k+1}+\frac{1}{(2k+1)(2k+3)}
    =\frac{k+1}{2k+1}

    ~~~~~~~~~~~~~~~~~~~

    Second, on the problem:

    2+7+...+(5n-3)=\frac{n}{2}(5n-1)

    I get to:

    P_k+(5k+2)
    \frac{k}{2}(5k+1)+(5k+2)

    And once again, I get stuck trying to finish the problem off. I know how to get to this point, I just fail to see how to finish it off

    Any and all help (and verification for the first problem) will be greatly appreciated.
    For the first problem, you did the induction correctly, except note that \frac{k(2k + 3)}{2k + 1)(2k + 3)} + \frac{1}{(2k + 1)(2k + 3)} = \frac{2k^2 + 3k + 1}{(2k + 1)(2k + 3)} = \frac{(2k + 1)(k + 1)}{(2k + 1)(2k + 3)} = \frac{k + 1}{2k + 3}, and that is the answer you are looking for. I assume you just made a typo.

    For the second problem, first note that the expression to be simplified is \frac{k}{2}(5k - 1) + 5k + 2, not \frac{k}{2}(5k+1)+(5k+2). Try simplifying it as \frac{1}{2}(5k^2 - k + 10k + 4) and factor.
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  3. #3
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    Quote Originally Posted by icemanfan View Post
    For the first problem, you did the induction correctly, except note that \frac{k(2k + 3)}{2k + 1)(2k + 3)} + \frac{1}{(2k + 1)(2k + 3)} = \frac{2k^2 + 3k + 1}{(2k + 1)(2k + 3)} = \frac{(2k + 1)(k + 1)}{(2k + 1)(2k + 3)} = \frac{k + 1}{2k + 3}, and that is the answer you are looking for. I assume you just made a typo.
    No typo. That's how I worked it out.

    For the second problem, first note that the expression to be simplified is \frac{k}{2}(5k - 1) + 5k + 2, not \frac{k}{2}(5k+1)+(5k+2). Try simplifying it as \frac{1}{2}(5k^2 - k + 10k + 4) and factor.
    Thanks for the assistance!
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