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Show that 1^2+3^2+ (2n+1)^2

  1. Apr 27, 2012 #1
    Show that 1^2+3^2+....(2n+1)^2

    1. The problem statement, all variables and given/known data

    show

    1^2+3^2+....+(2n+1)^2=1/3(n+1)(2n+1)(2n+3) whenever n is an element of natural numbers.



    2. Relevant equations



    3. The attempt at a solution

    So I started with the basis step of proving true for n=1

    so 1^2+.....+(2(1)+1)^2=1/3(1+1)(2+1)(2+3)

    so 10 = 1/3(2)(3)(5) =10 so true for n=1

    Assume true for n=k

    Show it true for n=k+1

    so 1^2+3^2+....(2k+1)^2+(2k+3)= 1/3(n+1)(2n+1)(2n+3)+(2k+3)^2

    as 1/3(n+1)(2n+1)(2n+3)=1^2+3^2+....(2k+1)^2

    But I don't know where to go from here??

    I have to get it in the form

    1/3(odd integer)(odd integer)(odd integer). dont I??
     
  2. jcsd
  3. Apr 27, 2012 #2

    tiny-tim

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    hi charmedbeauty! :smile:

    (try using the X2 button just above the Reply box :wink:)
    subtract? :wink:
     
  4. Apr 27, 2012 #3
    Re: Show that 1^2+3^2+....(2n+1)^2

    but what can I subtract from this

    1/3(k+1)(2k+1)(2k+3)+(2k+3)2

    if I expand out I get

    1/3(4k3+12k2+11k+3)+1/3(12k2+36k+27)

    rounding up terms..

    1/3(4k3+24k2+47k+30)

    where does the subtraction come from? I thought the two original terms above were separated by addition?
     
  5. Apr 27, 2012 #4

    tiny-tim

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    it'll help to factor out (2k+3) first :wink:
     
  6. Apr 27, 2012 #5
    Re: Show that 1^2+3^2+....(2n+1)^2

    you mean something like this..

    (2k+3)(1/3(k+1)(2k+1)+(2k+3))

    But I still cant see what to do?
     
  7. Apr 27, 2012 #6

    Dick

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    Re: Show that 1^2+3^2+....(2n+1)^2

    You want to show that result is the same as putting n to be n+1 in (1/3)(n+1)(2n+1)(2n+3). That's (1/3)(n+2)(2n+3)(2n+5). Is it?
     
  8. Apr 27, 2012 #7
    Re: Show that 1^2+3^2+....(2n+1)^2

    Yes I can see but the problem I keep on having is I'm always left with that stupid addition sign, how do I make it go away?
     
  9. Apr 27, 2012 #8

    Dick

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    Re: Show that 1^2+3^2+....(2n+1)^2

    If you expand out (1/3)(n+2)(2n+3)(2n+5) do you get the same thing as (1/3)(n+1)(2n+1)(2n+3)+(2n+3)^2?
     
  10. Apr 27, 2012 #9

    Ray Vickson

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    Re: Show that 1^2+3^2+....(2n+1)^2

    Your problem notation did not make clear whether you were dealing with problem (1) or (2) below:
    (1) sum_{j <= 2n+1} j^2 (all numbers from 1 to 2n+1), or
    (2) sum_{j <= 2n+1, j ODD} j^2 (odd numbers only from 1 to 2n+1).

    RGV
     
  11. Apr 27, 2012 #10
    Re: Show that 1^2+3^2+....(2n+1)^2

    Yes it did. The first two terms were 1^2+3^2, and that's enough to imply the second, as the first's second term would be 2^2, not 3^2.
     
  12. Apr 27, 2012 #11

    Ray Vickson

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    Re: Show that 1^2+3^2+....(2n+1)^2

    You'r right. Sorry. I need my morning coffee.

    RGV
     
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