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Math Help - Mathematical induction

  1. #1
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    Mathematical induction

    Hi, the question I have is:

    a) Use mathematical induction to prove that

    \sum\frac{1}{r(r+2)}=\frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)} for n=1,2,3...

    b) Deduce that

    \sum\frac{1}{r(r+2)}<\frac{3}{4} for n=1,2,3...
    justify your answer.

    For a) I have the following, so far:

    Let p(n) be the variable proposition \sum\frac{1}{r(r+2)}=\frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}

    For n=1, the sum is
    \frac{1}{1(1+2)}=\frac{1}{3}

    The right hand side gives:
    \frac{3}{4}-\frac{2*1+3}{2(1+1)(1+2)}=\frac{1}{3}

    So p(1) is tue.

    Assuming p(k) is true, I get:
    \sum\frac{1}{r(r+2)}=\frac{1}{3}+\frac{1}{8}+\frac  {1}{15}+...+\frac{1}{k(k+2)}=\frac{3}{4}-\frac{2k+3}{2(k+1)(k+2)}

    next I tried to deduce that p(k+1) is true, that is:
    \sum\frac{1}{r(r+2)}=\frac{1}{3}+\frac{1}{8}+\frac  {1}{15}+...+\frac{1}{k(k+2)}+\frac{1}{(k+1)(k+3)}=  \frac{3}{4}-\frac{2(k+1)+3}{2(k+2)(k+3)}

    Is this right so far?

    Thanks for your help
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by cozza View Post
    Hi, the question I have is:

    a) Use mathematical induction to prove that

    \sum\frac{1}{r(r+2)}=\frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)} for n=1,2,3...

    b) Deduce that

    \sum\frac{1}{r(r+2)}<\frac{3}{4} for n=1,2,3...
    justify your answer.

    For a) I have the following, so far:

    Let p(n) be the variable proposition \sum\frac{1}{r(r+2)}=\frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}

    For n=1, the sum is
    \frac{1}{1(1+2)}=\frac{1}{3}

    The right hand side gives:
    \frac{3}{4}-\frac{2*1+3}{2(1+1)(1+2)}=\frac{1}{3}

    So p(1) is tue.

    Assuming p(k) is true, I get:
    \sum\frac{1}{r(r+2)}=\frac{1}{3}+\frac{1}{8}+\frac  {1}{15}+...+\frac{1}{k(k+2)}=\frac{3}{4}-\frac{2k+3}{2(k+1)(k+2)}

    next I tried to deduce that p(k+1) is true, that is:
    \sum\frac{1}{r(r+2)}=\frac{1}{3}+\frac{1}{8}+\frac  {1}{15}+...+\frac{1}{k(k+2)}+\frac{1}{(k+1)(k+3)}=  \frac{3}{4}-\frac{2(k+1)+3}{2(k+2)(k+3)}

    Is this right so far?

    Thanks for your help
    yes

    just now add that  \displaystyle \frac {1}{(k+1)(k+3)} to the  \displaystyle \frac {3}{4} -\frac {2k+3}{2(k+1)(k+2) }
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  3. #3
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    Yes, you are summing from r=1 to r=n.
    fine so far.
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  4. #4
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    Quote Originally Posted by yeKciM View Post
    yes

    just now add that  \displaystyle \frac {1}{(k+1)(k+3)} to the  \displaystyle \frac {3}{4} -\frac {2k+3}{2(k+1)(k+2) }

    I am struggling with the algebra involved in doing the actual sum:

    \frac{3}{4}-\frac{2k+3}{2(k+1)(k+2)}+\frac{1}{(k+1)(k+3)}

    Are you able to give me any pointers?
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  5. #5
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    Quote Originally Posted by cozza View Post
    I am struggling with the algebra involved in doing the actual sum:

    \frac{3}{4}-\frac{2k+3}{2(k+1)(k+2)}+\frac{1}{(k+1)(k+3)}

    Are you able to give me any pointers?
    Your Proposition P(k) is

    \displaystyle\sum_{r=1}^k\frac{1}{r(r+2)}=\frac{3}  {4}-\frac{2k+3}{2(k+1)(k+2)}


    Your Proposition P(k+1) is

    \displaystyle\sum_{r=1}^{k+1}\frac{1}{r(r+2)}=\fra  c{3}{4}-\frac{2(k+1)+3}{2(k+2)(k+3)}

    If you can show that P(k+1) will be definately true if P(k) is true,
    then you will have completed the inductive step.

    We use \displaystyle\sum_{r=1}^{k+1}\frac{1}{r(r+2)}=\sum  _{r=1}^k\frac{1}{r(r+2)}+\frac{1}{(k+1)(k+3)}

    If P(k) is true, then we need to check if

    \displaystyle\frac{3}{4}-\frac{2k+3}{2(k+1)(k+2)}+\frac{1}{(k+1)(k+3)}=\fra  c{3}{4}-\frac{2(k+1)+3}{2(k+2)(k+3)}

    The fraction 3/4 is irrelevant to comparing the LHS and RHS of the immediately above,
    hence we want to know if

    \displaystyle\frac{1}{(k+1)(k+3)}-\frac{2k+3}{2(k+1)(k+2)}=-\frac{2k+5}{2(k+2)(k+3)}

    or

    \displaystyle\frac{2k+3}{2(k+1)(k+2)}-\frac{2}{2(k+1)(k+3)}=\frac{2k+5}{2(k+2)(k+3)}

    One way to check if the LHS equals the RHS is to multiply the 1st term of the LHS by \displaystyle\frac{k+3}{k+3}

    and multiply the 2nd term of the LHS by \displaystyle\frac{k+2}{k+2}

    and then multiply both sides (RHS and LHS) by (k+1).
    Then both sides will have a common denominator, so you only need compare numerators.

    If you prefer, you could post the steps you've taken and we can check your work.
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