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Induction question

  1. Jan 12, 2005 #1
    Hello, I am wondering how explicitly a result needs to be shown in induction questions.

    For example prove that 1² + 2² +....+n² = (1/6)n(n+1)(2n+1) where n is a natural number. Here is what I did.

    1. P(n) is the statement that the sum of the squares of the first n positive integers is (1/6)n(n+1)(2n+1). Show P(2) is true.

    Edit: Fixed first sentence of the above paragraph.

    2. Assume P(k) is true where k is a positive integer.

    [tex]
    1^2 + 2^2 + ... + k^2 = \frac{1}{6}k\left( {k + 1} \right)\left( {2k + 1} \right)
    [/tex]


    3. Show that P(k+1) is true.

    [tex]
    1^2 + 2^2 + ... + k^2 + \left( {k + 1} \right)^2 = \frac{1}{6}k\left( {k + 1} \right)\left( {2k + 1} \right) + \left( {k + 1} \right)^2
    [/tex]

    [tex]
    = \left( {k + 1} \right)\left[ {\frac{1}{6}k\left( {2k + 1} \right) + \left( {k + 1} \right)} \right]
    [/tex]

    [tex]
    = \frac{1}{6}\left( {k + 1} \right)\left[ {k\left( {2k + 1} \right) + 6\left( {k + 1} \right)} \right]
    [/tex]

    [tex]
    = \frac{1}{6}\left( {k + 1} \right)\left[ {2k^2 + 7k + 6} \right]
    [/tex]

    [tex]
    = \frac{1}{6}\left( {k + 1} \right)\left( {k + 2} \right)\left( {2k + 3} \right)
    [/tex]

    Is it fine up to the point above? Or do I need to also have the following?

    [tex]
    = \frac{1}{6}\left( {k + 1} \right)\left( {\left( {k + 1} \right) + 1} \right)\left( {2\left( {k + 1} \right) + 1} \right)
    [/tex]

    Any help appreciated.
     
    Last edited: Jan 13, 2005
  2. jcsd
  3. Jan 13, 2005 #2
    your last step is absolutely nessisory (sry, dunno how to spell this word)... The reason is: you must show in the P(n+1) case, your formulas is the exactly the same as P(n) with n replaced by (n+1)....
     
  4. Jan 13, 2005 #3
    first you must show it is true when n = 1

    which is easy enough, 1 = (1/6)*1*2*3

    then do n+1 as you did

    I personally would write that last line you wrote to make it very clear that n was replaced with K+1
     
  5. Jan 13, 2005 #4
    Cool, thanks for the help vincent and spectre.
     
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