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Trend in Square roots question [Curious Math Newb question]

  1. Jan 19, 2012 #1
    I noticed that 2 grows by 2 when it is squared, and 3 grows by 6 when it is squared, and 4 grows by 12, and 5 grows by 20... etc. etc.

    So 3's increase when squared is 4 more than 2's increase when squared, and 4's increase when squared is 6 greater than 3's increase when squared, and 5's increase when squared is 8 bigger than 4's......... etc. etc... <---The change from 4-5 is 2 bigger than the change from 3-4, and so on..

    Why does this trend exist? Does it mean anything?

    Thanks guys
    Last edited: Jan 19, 2012
  2. jcsd
  3. Jan 19, 2012 #2

    Simon Bridge

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    Well lets see...
    It helps to be able to write number relations as math vis:
    the increase in a number n when you square it is given by - [itex]I_n=n^2-n[/itex]

    the difference in the adjacent increases would be: [itex]D_n=I_n-I_{n-1}[/itex] which would be:[tex]D_n=\big [ n^2-n \big ] - \big [ (n-1)^2 - (n-1) \big ] = 2(n-1)[/tex]... which is linear.

    That what you mean?
  4. Jan 19, 2012 #3
    Oh thanks that was super helpful

    The second part of my post was trying to say Dn - Dn-1 = 2 and that can be found by 2(n-1) - 2(n-1-1) = 2!
  5. Jan 19, 2012 #4
    Notice also the same linearity in the simpler progression 9 - 4 = 5, 16 - 9 = 7, 25 - 16 = 9 ...

    If you grasp visual concepts more easily, look at this:


    The multicoloured square on the left represents your original series, that on the right represents the simpler one mentioned above. In each case the two red squares are the difference between successive terms (in this case 62 and 72) - notice that in each large square the green and blue L-shapes are congruent.
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