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Math Help - [SOLVED] Prove using mathematical induction

  1. #1
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    [SOLVED] Prove using mathematical induction

    I need some help with mathematical induction. It's driving me insane.

    Use mathematical induction to prove that:
    1/3 + 2/3^2 + 3/3^3 +...+ n/3^n = 1/4 (3 - 2n + 3/ 3^n )

    Basis step: Prove 1 exists in S.

    n = 1 LHS = 1/3^1 = 1/3

    RHS = 1/4 (3 - 2(1) + 3/ 3^1 ) = 1/3

    So 1 exists in S

    Inductive step:

    Assume k exists in S. We must prove k+1 exists in S

    So assume 1/3 + 2/3^2 + 3/3^3 +...+ k/3^k = 1/4 (3 - 2k + 3/ 3^k )

    LHS = 1/3 + 2/3^2 + 3/3^3 +...+ k/3^k + (k + 1 /3^K+1)

    RHS =  1/4 (3 - 2k + 3/ 3^k ) + (k + 1 /3^K+1)

    After this i just get lost, i think i have a problem when trying to perform arithmetic on it. Any help would be much appreciated.

    I also have another question:
    Find a fourth order polynomial with integer coefficients for which 1 + 2root3 - root7 is a root.

    SO we have x = 1 + 2root3 - root7

    x - 1 =2root3 - root7

    (x - 1)^2 =12 - 7 + root21

    (x^2 - 2x - 3)^2 = 21

    (x^4 - 2x^2 +12 = 0


    any help would be much appreciated!
    Last edited by sterps; March 19th 2010 at 11:00 PM.
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  2. #2
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    Quote Originally Posted by sterps View Post
    I need some help with mathematical induction. It's driving me insane.

    Use mathematical induction to prove that:
    1/3 + 2/3^2 + 3/3^3 +...+ n/3^n = 1/4 (3 - 2n + 3/ 3^n )

    Basis step: Prove 1 exists in S.

    n = 1 LHS = 1/3^1 = 1/3

    RHS = 1/4 (3 - 2(1) + 3/ 3^1 ) = 1/3

    So 1 exists in S

    Inductive step:

    Assume k exists in S. We must prove k+1 exists in S

    So assume 1/3 + 2/3^2 + 3/3^3 +...+ k/3^k = 1/4 (3 - 2k + 3/ 3^k )

    LHS = 1/3 + 2/3^2 + 3/3^3 +...+ k/3^k + (k + 1 /3^K+1)

    RHS =  1/4 (3 - 2k + 3/ 3^k ) + (k + 1 /3^K+1)

    After this i just get lost, i think i have a problem when trying to perform arithmetic on it. Any help would be much appreciated.

    I also have another question:
    Find a fourth order polynomial with integer coefficients for which 1 + 2root3 - root7 is a root.

    SO we have x = 1 + 2root3 - root7

    x - 1 =2root3 - root7

    (x - 1)^2 =12 - 7 + root21

    (x^2 - 2x - 3)^2 = 21

    (x^4 - 2x^2 +12 = 0


    any help would be much appreciated!

    Read here: http://www.mathhelpforum.com/math-he...-question.html

    Tonio
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  3. #3
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    For the second part of your question, you've got the right idea, but you need to be careful doing the algebra:

    x=1+2\sqrt{3}-\sqrt{7}

    x-1=2\sqrt{3}-\sqrt{7}

    (x-1)^2=4*3-4*\sqrt{3}*\sqrt{7}+7=19-4\sqrt{21}

    x^2-2x+1=19-4\sqrt{21}

    x^2-2x-18=-4\sqrt{21}

    (x^2-2x-18)^2=16*21=336

    x^4-4x^3-32x^2+72x+324=336

    x^4-4x^3-32x^2+72x-12=0

    A tip for these boards: don't ask two questions in a thread. Someone comes by and answers one of your questions (as Tonio did), and then it looks like your question is answered, so no one sees your second question. There's no problem with starting two threads or even more, especially with two very different questions like you had in this thread.

    - Hollywood
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  4. #4
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    Thanks for the replies and the help. I ended up figuring out the 2nd question.
    Thanks to both tonio and you hollywood for your help. Ill take your advice into account next time.
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