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: Finding the sum of 1^3 + 2^3 + + n^3 by induction

  1. Jul 6, 2014 #1
    [tex]1^3+2^3+...+n^3 = \left[ \frac{n(n+1)}{2}\right]^2; n\ge 1[/tex]

    [tex]P(1) = 1^3 = \frac{8}{8} = 1[/tex]

    [tex]P(k) = 1^3+...+k^3 = \left[ \frac{k(k+1)}{2}\right]^2[/tex] (induction hypothesis)

    [tex]P(k+1) = 1^3+...+k^3+(k+1)^3 = \left[\frac{(k+1)(k+2)}{2}\right]^2[/tex]

    I start getting stuck here
    I foiled it out then let m = P(k)

    [tex]\left[ m + \frac{2(k+1)}{2}\right]^2[/tex]
     
    Last edited: Jul 6, 2014
  2. jcsd
  3. Jul 6, 2014 #2
    You probably meant ##m = \sqrt{P(k)}##. Now expand the square and use the induction hypothesis.
     
  4. Jul 6, 2014 #3

    ehild

    User Avatar
    Homework Helper

    You assume that [tex]P(k) = 1^3+...+k^3 = \left[ \frac{k(k+1)}{2}\right]^2[/tex] To get P(k+1), you have to add (k+1)^3 to P(k). [tex]P(k+1)=P(k)+(k+1)^3=\left[ \frac{k(k+1)}{2}\right]^2+(k+1)^3[/tex] Write it out, and show that it is equal to [tex]P(k+1) = \left[ \frac{(k+1)(k+2)}{2}\right]^2[/tex]
    pull out (k+1)2

    ehild
     
  5. Jul 6, 2014 #4

    [tex]\left[\frac{(k+1)(k+2)}{2}\right]^2 = \left[ \frac{k(k+1)}{2}\right]^2+(k+1)^3[/tex]

    [tex] (k+1)^2 \left[\frac{(k+2)}{2}\right]^2=(k+1)^2\left[\frac{k^2}{2^2} + (k+1)\right][/tex]

    [tex] (k+1)^2 \left[\frac{(k+2)}{2}\right]^2=(k+1)^2\left[\frac{k^2+4k+4}{2^2}\right][/tex]

    [tex] (k+1)^2 \left[\frac{(k+2)}{2}\right]^2=(k+1)^2\left[\frac{(k+2)^2}{2^2}\right][/tex]

    ???
     
  6. Jul 6, 2014 #5

    ehild

    User Avatar
    Homework Helper

    Apply that a2b2=(ab)2

    [tex](k+1)^2 \left[\frac{(k+2)}{2}\right]^2=\left[(k+1)\frac{(k+2)}{2}\right]^2= P(k+1)[/tex]

    is it familiar? Is it what you wanted to arrive at?


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

Have something to add?
Draft saved Draft deleted