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Math Help - Induction

  1. #1
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    Induction

    Verify by induction that the solution of recurrence:

    <br />
F(1) = 1<br />
    F(n) = F(n-1) +3n+2

    is F(n) = \frac{(3n^2)}{2} + \frac{7n}{2-4}
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  2. #2
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    Quote Originally Posted by Apprentice123 View Post
    Verify by induction that the solution of recurrence:

    <br />
F(1) = 1<br />
    F(n) = F(n-1) +3n+2

    is F(n) = \frac{(3n^2)}{2} + \frac{7n}{2-4}
    2- 4= -2 do you mean F(n)= \frac{3n^2}{2}- \frac{7n}{2}= \frac{3n^2- 7n}{2}?
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    2- 4= -2 do you mean F(n)= \frac{3n^2}{2}- \frac{7n}{2}= \frac{3n^2- 7n}{2}?
    Sory the correct is F(n) = \frac{3n^2}{2} + \frac{7n}{2} -4
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  4. #4
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    Quote Originally Posted by Apprentice123 View Post
    Sory the correct is F(n) = \frac{3n^2}{2} + \frac{7n}{2} -4
    While it can be difficult to solve equations, it can be much easier to verify a solution. With F(n)= \frac{3n^2}{2}+ \frac{7n}{2}- 4, F(n-1)= \frac{3(n-1)^2}{2}+ \frac{7(n-1)}{2}- 4= \frac{3n^2- 6n+ 3+ 7n- 7- 8}{2}= \frac{3n^2+ n- 12}{2}.

    Then F(n-1)+ 3n+ 2= \frac{3n^2+ n- 12}{2}+ 3n+ 2 = \frac{3n^2+ n- 12}+ \frac{6n+ 4}{2}= \frac{3n^2+ 7n- 8}{2}= \frac{3n^3}{2}+ \frac{7n}{2}- 4= F(n).
    Last edited by HallsofIvy; May 7th 2009 at 03:41 AM.
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    Thank you
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