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Induction Question

  1. Dec 6, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex]\sum[/itex] i=1 to n[itex]\sqrt{1+(1/i^2)+(1/(1+i)^2)}[/itex] = n(n+2)/n+1

    2. The attempt at a solution

    First I did the base case of p(1) showing 3/2 on the LHS equals the 3/2 on the RHS.
    Then I assumed p(k) and wrote out the formula with k in it.
    Then prove p(k+1)= p(k)+ [itex]\sqrt{1+1/(k+1)^2+1/(k+2)^2}[/itex]
    =k(k+2)/k+1 + [itex]\sqrt{1+1/(k+1)^2+1/(k+2)^2}[/itex]
    Then I squared each to get rid of the square root.
    (k(k+2)/(k+1))^2+ (k+1)^2/(k+1)^2 + 1/(k+1)^2 + 1/(k+2)^2
    Then I factored everything out
    ((k^4+4k^3+4k)/(k+1)^2) + ((k^2+2k+1)/(k+1)^2) + 1/(k+1)^2 + 1/(k+2)^2.
    Basically I'm having a brain freeze on how to get (k+2) as the common denominator. Any Guidance would be great thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 6, 2012 #2

    micromass

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    So what you basically did was

    [tex](a+b)^2=a^2+b^2[/tex]

    This formula is not valid!!! And this is also what causes your mistake.
     
  4. Dec 6, 2012 #3
    Sorry I'm not following.. I understand what you're saying I did, just confused on which part I did the (a+b)^2 = a^2 + b^2
     
  5. Dec 6, 2012 #4

    micromass

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    Correct me if I'm wrong, but you wanted to square

    [tex]\frac{k(k+2)}{k+1} + \sqrt{1+\frac{1}{(k+1)^2}+\frac{1}{(k+2)^2}}[/tex]

    So it seems to me that you did the following:

    [tex]
    \begin{eqnarray*}
    \left(\frac{k(k+2)}{k+1} + \sqrt{1+\frac{1}{(k+1)^2}+\frac{1}{(k+2)^2}}\right)^2
    & = & \left(\frac{k(k+2)}{k+1}\right)^2 + \left(\sqrt{1+\frac{1}{(k+1)^2}+\frac{1}{(k+2)^2}}\right)^2\\
    & = & \frac{k^2(k+2)^2}{(k+1)^2} + 1+\frac{1}{(k+1)^2}+\frac{1}{(k+2)^2}\\
    & = & \frac{k^2(k+2)^2}{(k+1)^2} + \frac{(k+1)^2}{(k+1)^2}+\frac{1}{(k+1)^2}+\frac{1}{(k+2)^2}
    \end{eqnarray*}
    [/tex]

    Is this what you you were trying to do?? In that case, it is the first equality that is not correct.
     
  6. Dec 6, 2012 #5
    Sorry I had it written down on my paper but I didn't type it right! I'm just having a brain freeze on what I need to multiply the other rationals to have a denominator of (k+2)^2. Assuming that's the correct next step.
     
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