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Math Help - Induction

  1. #1
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    Induction

    Hi Guys I have a question for you.

    The statement is :
    Let ao = 3, and let a_(n+1) = (an + 7)^(1/2) if n > 0. Prove that 3 < a < 4
    for all n > 0.


    Thanks a lot for the help.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by joksy View Post
    Hi Guys I have a question for you.

    The statement is :
    Let ao = 3, and let a_(n+1) = (an + 7)^(1/2) if n > 0. Prove that 3 < a < 4
    for all n > 0.


    Thanks a lot for the help.
    prove that 3 < a,, < 4 ??

    what is a,, ?
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  3. #3
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    Hello, joksy!

    Let a_0 = 3, and let a_{n+1} \:=\:\sqrt{a_n + 7}

    Prove that: . 3 < a_n < 4,\:\text{ for all }n > 0

    Verify S(1)\!:\;\;a_1 \:=\:\sqrt{a_0+7} \:=\:\sqrt{3 + 7} \:\approx\:3.16 . . . True


    Assume S(k)\!:\;\;3 \:<\:a_k \:<\:4

    Add 7: . 10 \:<\:a_k + 7 \:<\:11

    \text{Take square roots: }\;\sqrt{10} \:<\:\underbrace{\sqrt{a_k+7}}_{\text{This is }a_{k+1}} \:<\:\sqrt{11}

    So we have: .  3.162 \:<\:a_{k+1} \:<\:3.317 \quad\hdots\quad 3 \:<\:a_{k+1} \:<\:4


    We have shown that S(k+1) is true.
    . . The inductive proof is complete.

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  4. #4
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    doh. Thanks a lot jhevon, I did see this trick... I have to work a lot on more exercise !! But anyway thanks a lot
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by joksy View Post
    doh. Thanks a lot jhevon, I did see this trick... I have to work a lot on more exercise !! But anyway thanks a lot
    well, it was Soroban that helped you
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  6. #6
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    Sorry where a red your name... so THANKS SOROBAN and sorry for the misunderstanding. and for you thanks for tell me the error.


    Have a nice day guys
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