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Math Help - a question about induction

  1. #1
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    a question about induction

    prove that sigma(1/r(r+1)(r+2)=0.25-1/2(n+1)(n+2)
    it is the proving for k+1 i have a question about, do you put both sides equal to each other and the expand/simplify one side using the fact that you have assumed it is true for k and the show that both sides equal each other. my teacher went over this recently however i belive she got it wrong, any help would be greatly appreciated.
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  2. #2
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    \sum\limits_{r=1}^{n}{\frac{1}{r(r+1)(r+2)}}=\frac  {1}{4}-\frac{1}{2(n+1)(n+2)}.

    Verify that n=1 is true & assume valid your proposition for n=k. Hence, we must prove that \sum\limits_{r=1}^{k+1}{\frac{1}{r(r+1)(r+2)}}=\fr  ac{1}{4}-\frac{1}{2(k+2)(k+3)}.

    Observe that,

    \begin{aligned}<br />
   \sum\limits_{r=1}^{k+1}{\frac{1}{r(r+1)(r+2)}}&=\s  um\limits_{r=1}^{k}{\frac{1}{r(r+1)(r+2)}}+\sum\li  mits_{r=k+1}^{k+1}{\frac{1}{r(r+1)(r+2)}} \\ <br />
 & =\frac{1}{4}-\frac{1}{2(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)} \\ <br />
 & =\frac{1}{4}-\frac{1}{(k+1)(k+2)}\left( \frac{1}{2}-\frac{1}{k+3} \right) \\ <br />
 & =\frac{1}{4}-\frac{1}{(k+1)(k+2)}\cdot \frac{k+1}{2(k+3)},<br />
\end{aligned}

    which is after simplification, the result we wanted to prove. \blacksquare
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  3. #3
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    it is this simplification which i am not sure about. must we rearrange that to the form 0.25-1/2(n+1)(n+2) or can we simply set it equal to that and cancel it down?? thanks
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    You're proving the statement by mathematical induction. You're assuming your proposition valid for n=1 and for some n=k. The next step is to prove that for n=k+1 the equality does hold, which is I've done above.

    If you want to "rearrange" to get \frac{1}{4}-\frac{1}{2(n+1)(n+2)}, you'll need no induction, but telescoping series will be helpful.
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