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Induction Help

  1. Dec 10, 2012 #1
    1. The problem statement, all variables and given/known data

    ∑ i=1 to n1+(1/i2)+(1/(1+i)2)−−−−−−−−−−−−−−−−−−−−√ = 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)+ 1+1/(k+1)2+1/(k+2)2−−−−−−−−−−−−−−−−−−−−−−√
    =k(k+2)/k+1 + 1+1/(k+1)2+1/(k+2)2−−−−−−−−−−−−−−−−−−−−−−√
    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
    Now I'm stuck any Guidance would be great thanks!
     
  2. jcsd
  3. Dec 10, 2012 #2

    SammyS

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    The above line is nonsensical. Not enough parentheses, probably missing exponentiation, dangling radical ...

    Do you mean [itex]\displaystyle \sum_{i=1}^{n}\sqrt{1+1/(i^2)+1/(1+i)^2}\ \ = \ \ n(n+2)/(n+1)\ ?[/itex]

    If so, then I must be a much better mind reader than I have ever been given credit for !
     
  4. Dec 10, 2012 #3
    yes sorry about that!
     
  5. Dec 10, 2012 #4

    SammyS

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    You answered "yes" to what? Please be precise .
     
  6. Dec 10, 2012 #5
    Yes I meant that!
     
    Last edited by a moderator: Dec 10, 2012
  7. Dec 10, 2012 #6

    SammyS

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    As you stated: For the induction step, you assume that
    [itex]\displaystyle \sum_{i=1}^{k}\sqrt{1+\frac{1}{i^2}+\frac{1}{(1+i)^2}}\ = \ \frac{k(k+2)}{k+1}[/itex]​
    for k ≥ 1 . From that assumption, show that it follows that
    [itex]\displaystyle \sum_{i=1}^{k+1}\sqrt{1+\frac{1}{i^2}+\frac{1}{(1+i)^2}}\ = \ \frac{(k+1)(k+3)}{k+2}\ .[/itex]​

    So look at [itex]\displaystyle \ \ \sum_{i=1}^{k+1}\sqrt{1+\frac{1}{i^2}+\frac{1}{(1+i)^2}}\,,\ \ [/itex] which you can write as [itex]\displaystyle \ \ \sqrt{1+\frac{1}{(k+1)^2}+\frac{1}{(k+2)^2}}+\sum_{i=1}^{k}\sqrt{1+\frac{1}{i^2}+\frac{1}{(1+i)^2}}\ .[/itex]

    Now use you assumption, then do some algebra.
     
  8. Dec 11, 2012 #7
    I'm just having a trouble showing, [itex]\frac{k(k+2)}{k+1} + \sqrt{1+\frac{1}{(k+1)^2}+\frac{1}{(k+2)^2}}\ .[/itex] = [itex]\frac{k+1(k+1)}{k+1+2}\ .[/itex]
     
  9. Dec 11, 2012 #8
    I'm just having a trouble showing, [itex]\frac{k(k+2)}{k+1} + \sqrt{1+\frac{1}{(k+1)^2}+\frac{1}{(k+2)^2}}+\sum_{i=1}^{k}\sqrt{1+\frac{1}{i^2}+\frac{1}{(1+i)^2}}\ .[/itex] = [itex]\frac{k+1(k+1+2)}{k+1}\ .[itex]
     
  10. Dec 11, 2012 #9
    I'm just having a trouble showing, [itex]\frac{k(k+2)}{k+1} + \sqrt{1+\frac{1}{(k+1)^2}+\frac{1}{(k+2)^2}}\ [/itex] = [itex]\frac{k+1(k+1+2)}{k+1}\ [/itex]
     
  11. Dec 11, 2012 #10
    What you really need to show next is
    $$
    \frac{k(k+2)}{k+1}+\sqrt{1+\frac1{(k+1)^2}+\frac1{(k+2)^2}} = \frac{\color{red}{(k+1)} (k+3)}{k+\color{red}2}.
    $$
    I'd try subtracting ##\frac{k(k+2)}{k+1}## on both sides and then squaring the results.
     
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