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Math Help - Proof: recursive and closed formulas

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    Proof: recursive and closed formulas

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    Last edited by mathh18; February 28th 2010 at 05:51 PM.
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    Quote Originally Posted by mathh18 View Post
    i'm trying to prove that recursive formula 2[a(sub)k-1] -3 and closed formula [a(sub)n]= 2^(n-1) +3 are equal.
    i have [a(sub)m] = 2[a(sub)m-1]-3 =
    2(2^(m-3)+3) -3 =
    2^(m-2) +3
    that is close, but slightly off than what i'm trying to prove..what should i do to make it 2^(m-1) +3 ?

    sorry for the typing, it's confusing on the computer
    You can learn to use LaTeX.
    [tex]a_n =2^{n-1}-3[/tex] gives a_n =2^{n-1}-3.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mathh18 View Post
    i'm trying to prove that recursive formula 2[a(sub)k-1] -3 and closed formula [a(sub)n]= 2^(n-1) +3 are equal.
    i have
    [a(sub)m] = 2[a(sub)m-1]-3 =
    2(2^(m-3)+3) -3 =
    2^(m-2) +3
    that is close, but slightly off than what i'm trying to prove..what should i do to make it 2^(m-1) +3 ?

    sorry for the typing, it's confusing on the computer
    I think this says prove that a_n=2a_{n-1}+3\implies a_n=2^{n-1}+3. Know induction?
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    sort of..i'm in the process of learning it. how does it work?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mathh18 View Post
    sort of..i'm in the process of learning it. how does it work?
    Your English teachers lied, this is your bff
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    ..
    Last edited by mathh18; February 28th 2010 at 05:51 PM.
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mathh18 View Post
    haha yeah! that's what i'm doing. i made a chart for n=1 and n=2 n=m-1 to show its true for all of those values. then i tried proving it for m and got stuck
    Much simpler than that. Note that if we assume that a_{n}=2^{n-1}+3 then a_{n+1}=2a_n+3=2\left(2^{n-1}+3\right)-3=2^n+6-3=2^n+3
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  8. #8
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    thanks that makes sense. i also just found a really stupid error in my work.
    got it now!
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