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Math Help - Some help on mathematical induction?

  1. #1
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    Some help on mathematical induction?

    Have some trouble completing these questions:
    Any help will be greatly appreciated:

    1 Use mathematical induction to prove that 5^n+9^n+2 is divisible by 4 for all n>0

    2a) use mathematical induction to prove that:
    1/(2r-1)(2r+1)=n/2n+1 for all n>0
    b) Hence show that the sum of the first (n +1) terms of the series
    1/3+1/15+1/35+1/63+....is (n+1)/(2n+3)

    3) (1)(1!)+(2)(2!)+(3)(3!)+...+(n)(n!)=(n+1 where n>0 Find the minimum number of terms of the series for the sum to exceed 10^9
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    1.) Base case of  n=0 :  5^0+9^0+2 = 4 .

    Now suppose  5^n+9^n+2 is divisible by  4 , then  5^n+9^n+2 = 4k for some natural number  k .

    Then  4 \cdot 5^n+8 \cdot 9^n+5^n+9^n+2 = 4 \cdot 5^n+8 \cdot 9^n+4k .

    Upon simplification we get  5^{n+1}+9^{n+1}+2 = 4(k+5^n+2\cdot9^n) = 4k' . And we're done.
    Last edited by chiph588@; November 29th 2009 at 06:15 PM.
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by blaze View Post
    2a) use mathematical induction to prove that:
    1/(2r-1)(2r+1)=n/2n+1 for all n>0
    b) Hence show that the sum of the first (n +1) terms of the series
    1/3+1/15+1/35+1/63+....is (n+1)/(2n+3)
    What is  r here?
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  4. #4
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    Quote Originally Posted by chiph588@ View Post
    What is  r here?
    Sigma Notation: n (on top) r=1 (on bottom)
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  5. #5
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    Quote Originally Posted by blaze View Post
    Have some trouble completing these questions:
    Any help will be greatly appreciated:

    1 Use mathematical induction to prove that 5^n+9^n+2 is divisible by 4 for all n>0

    2a) use mathematical induction to prove that:
    1/(2r-1)(2r+1)=n/2n+1 for all n>0
    b) Hence show that the sum of the first (n +1) terms of the series
    1/3+1/15+1/35+1/63+....is (n+1)/(2n+3)

    3) (1)(1!)+(2)(2!)+(3)(3!)+...+(n)(n!)=(n+1 where n>0 Find the minimum number of terms of the series for the sum to exceed 10^9
    So for 2) you're trying to prove

    \sum_{r = 1}^n\frac{1}{(2r - 1)(2r + 1)} = \frac{n}{2n + 1} for all n > 0?


    Base step: n = 1.

    LHS = \sum_{r = 1}^n\frac{1}{(2r - 1)(2r + 1)}

     = \frac{1}{3}.


    RHS = \frac{1}{3}

     = LHS as required.


    Inductive step: Assume the statement is true for n = k.

    So \sum_{r = 1}^k\frac{1}{(2r - 1)(2r + 1)} = \frac{k}{2k + 1}.


    We need to show that the equation holds true for n = k + 1.

    LHS = \sum_{r = 1}^{k + 1}\frac{1}{(2r - 1)(2r + 1)}

     = \sum_{r = 1}^k\frac{1}{(2r - 1)(2r + 1)} + \frac{1}{(2k + 1)(2k + 3)}

     = \frac{k}{2k + 1} + \frac{1}{(2k + 1)(2k + 3)}

     = \frac{k(2k + 3)}{(2k + 1)(2k + 3)} + \frac{1}{(2k + 1)(2k + 3)}

     = \frac{2k^2 + 3k + 1}{(2k + 1)(2k + 3)}

    = \frac{(2k + 1)(k + 1)}{(2k + 1)(2k + 3)}

     = \frac{k + 1}{2k + 3}.


    RHS= \frac{k + 1}{2k + 3}

     = LHS as required.


    Q.E.D.
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  6. #6
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    Quote Originally Posted by Prove It View Post
    So for 2) you're trying to prove

    \sum_{r = 1}^n\frac{1}{(2r - 1)(2r + 1)} = \frac{n}{2n + 1} for all n > 0?


    Base step: n = 1.

    LHS = \sum_{r = 1}^n\frac{1}{(2r - 1)(2r + 1)}

     = \frac{1}{3}.


    RHS = \frac{1}{3}

     = LHS as required.


    Inductive step: Assume the statement is true for n = k.

    So \sum_{r = 1}^k\frac{1}{(2r - 1)(2r + 1)} = \frac{k}{2k + 1}.


    We need to show that the equation holds true for n = k + 1.

    LHS = \sum_{r = 1}^{k + 1}\frac{1}{(2r - 1)(2r + 1)}

     = \sum_{r = 1}^k\frac{1}{(2r - 1)(2r + 1)} + \frac{1}{(2k + 1)(2k + 3)}

     = \frac{k}{2k + 1} + \frac{1}{(2k + 1)(2k + 3)}

     = \frac{k(2k + 3)}{(2k + 1)(2k + 3)} + \frac{1}{(2k + 1)(2k + 3)}

     = \frac{2k^2 + 3k + 1}{(2k + 1)(2k + 3)}

    = \frac{(2k + 1)(k + 1)}{(2k + 1)(2k + 3)}

     = \frac{k + 1}{2k + 3}.


    RHS= \frac{k + 1}{2k + 3}

     = LHS as required.


    Q.E.D.
    I already got the answer down- I need 2b) and 3) please
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  7. #7
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    Quote Originally Posted by blaze View Post
    I already got the answer down- I need 2b) and 3) please
    If you understand 2a then there is no way you didn't do 2b along the way. The proof of 2a by induction requires showing 2b. Please post one question per thread in the future and show what work you have attempted.

    So what do you need help with for 2?
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  8. #8
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    I don't get how you would pose the answer for 2b)

    As for 3), I have proved the mathematical induction theory but not sure how to " Find the minimum number of terms of the series for the sum to exceed 10^9"
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