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Math Help - factorials

  1. #1
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    factorials

    I am trying to show how to solve this using mathematical induction.
    I don't think I quite understand what to do with the factorials:

    Prove that 2x6x10...x(4n-2) = (2n)!/n!
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  2. #2
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    Hello, Nichelle14!

    Using mathematical induction, prove that: .  2 \cdot 6 \cdot 10\,\cdots\,(4n-2) \;= \;\frac{(2n)!}{n!}
    Verify S(1):\;\;2 \:=\:\frac{(2\cdot1)!}{1!} \:= \:2 . . . true!

    Assume S(k):\;\;2\cdot6\cdot10\,\cdots\,(4k-2)\;= \;\frac{(2k)!}{k!}


    Multiply both sides by 4(k+1) - 2 \:= \:4k + 2

    . . 2\cdot6\cdot10\,\hdots\,(4k-2)\cdot(4[k+1] - 2) \;= \;\frac{(2k)!}{k!}\cdot(4k + 2)


    The left side is the left side of S(k+1).

    We must show that the right side is: . \frac{(2k + 2)!}{(k+1)!} ... the right side of S(k+1).


    We have: . \frac{(2k)!\cdot2(2k+1)}{k!}

    Multiply top and bottom by k + 1:\;\;\frac{(2k)!\cdot2(2k+1)}{k!}\cdot\frac{k + 1}{k + 1}

    . . and we have: . \frac{(2k)!\cdot(2k+1)\cdot2(k+1)}{k!(k+1)} \;= \frac{(2k)!(2k+1)(2k+2)}{(k+1)!} \;= \;\frac{(2k+2)!}{(k+1)!}


    Therefore, we have proved S(k+1):

    . . 2\cdot6\cdot10\,\cdots\,(4k-2)\cdot(4k + 2) \;= \;\frac{(2k+2)!}{(k+1)!} . . . ta-DAA!

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  3. #3
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    thanks. I was doing it right, but got stumped until you showed to mulitply by (k+1)
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