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Math Help - Mathematical Induction

  1. #1
    Member helloying's Avatar
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    Mathematical Induction

    U_{n+1} = U_n +4+4(U_n+1)^{1/2}
    prove that U_n= 4(n+1)^2 -1
    Last edited by helloying; September 6th 2010 at 08:01 PM.
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  2. #2
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    Hello, helloying!

    There must be typos in the problem.
    The claim is not true.


    U_{n+1} \:=\:U_n +4+ 4(n+1)^{\frac{1}{2}} .[1]
    . . . . . . . . U_1 = 15

    Prove by induction that: . U_n \:=\:4(4n+1)^2 -1 .[2]

    The recursion formula has a square root.
    I don't see how the n^{th} term could always be an integer.

    In fact, we can't determine the second term of the sequence!


    In [1], let n = 1\!:\;\;U_2 \;=\;U_1 + 4 + 4\sqrt{1+1}

    . . . . . . . . . . . . . . . . =\; 15 + 4 + 4\sqrt{2}

    . . . . . . . . . . . . . . . . =\;19 + 4\sqrt{2}


    In [2], let n = 2:\;\;U_2 \;=\;4(4\cdot2+1)^2-1

    . . . . . . . . . . . . . . . . =\; 4(81) - 1

    . . . . . . . . . . . . . . . . =\; 323

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  3. #3
    Member helloying's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, helloying!

    There must be typos in the problem.
    The claim is not true.


    The recursion formula has a square root.
    I don't see how the n^{th} term could always be an integer.

    In fact, we can't determine the second term of the sequence!


    In [1], let n = 1\!:\;\;U_2 \;=\;U_1 + 4 + 4\sqrt{1+1}

    . . . . . . . . . . . . . . . . =\; 15 + 4 + 4\sqrt{2}

    . . . . . . . . . . . . . . . . =\;19 + 4\sqrt{2}


    In [2], let n = 2:\;\;U_2 \;=\;4(4\cdot2+1)^2-1

    . . . . . . . . . . . . . . . . =\; 4(81) - 1

    . . . . . . . . . . . . . . . . =\; 323
    Thanks i have edited my post. sorry about the typo
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  4. #4
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    1. Write the claim that you need to prove in the following form: "for all n >=1, ... ". Call the part denoted by ellipses, which contains n, by P(n). This is the induction statement, or the induction hypothesis (IH). Note that for each particular n, P(n) is not a number; it is a proposition that is either true or false.

    2. Prove P(1). This is the base case.

    3. Prove "for all n, P(n) implies P(n+1)". For this, fix an arbitrary n and assume P(n) (that's why it is called the IH). Use it to prove P(n+1).
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  5. #5
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    Quote Originally Posted by helloying View Post
    U_{n+1} = U_n +4+4(U_n+1)^{1/2}

    Prove\ that\ U_n= 4(n+1)^2 -1
    You can set it up as follows if you like...

    \displaystyle\ U_{k+1}=U_k+4+4\left(U_k+1\right)^{\frac{1}{2}}

    Then...

    P(k)

    U_k=4(k+1)^2-1


    P(k+1)

    U_{k+1}=4(k+2)^2-1

    For the proof, try to show

    U_k+4+4\left(U_k+1\right)^{\frac{1}{2}}=4(k+2)^2-1

    hence, try to show that

    \displaystyle\ [4(k+1)^2-1]+4+4[4(k+1)^2-1+1]^{\frac{1}{2}}=4(k+2)^2-1

    When you get through that, test for the first k or n.
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