1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
    So what you basically did was


    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
    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:

    \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}

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook