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Adding consecutive squares?

  1. Oct 2, 2005 #1
    ok, i need to derive a forumla that will add the consecutive squares of n numbers.

    for example [tex]1^2 + 2^2 + 3^2 + ... + (n-2)^2 + (n-1)^2 + (n)^2[/tex]

    I have worked on this problem for quite some time and havent been able to come up with anything.

    I do know that the sum of consecutive numbers starting at one is

    [tex]\frac{n}{2} (n+1)[/tex]

    A very detailed explanation would be excellent as that's what my professor wants.
    Last edited: Oct 2, 2005
  2. jcsd
  3. Oct 2, 2005 #2
    [tex]S(n) = \frac{n(n+1)(2n+1)}{6}[/tex].

    You should be able to prove it by induction.
  4. Oct 2, 2005 #3
    what's induction?

    can you explain how you came to the answer?
  5. Oct 2, 2005 #4
    inductions like this...
    prove it works for 1
    assume it works for n
    and prove it works for n+1
  6. Oct 2, 2005 #5
    s=summation of t
    we will get s
  7. Oct 2, 2005 #6
    Dr. Math has answered a lot of questions concerning the sum of consecutive squares here. He explains that there are several ways to derive the formula.
  8. Oct 2, 2005 #7
    ok, so i understand what inductions are, but can you explain how you got to the proof for the sum of the sequence of [tex]n^2[/tex]

    *** edit *** i just posted the above before reading the previous 3 posts. i'll go ahead and read dr math's explanation and then come back to this :)
  9. Oct 2, 2005 #8

    i need to derive the forumula... not prove it :)
  10. Oct 2, 2005 #9
    Sequence of Differences. Search here or Dr. Math, there are explanations at both places.
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