Results 1 to 7 of 7

Math Help - exer2

  1. #1
    Newbie
    Joined
    Aug 2008
    Posts
    21

    exer2

    Demonstrate that:
    \<br />
\left[ {\sqrt 1 } \right] + \left[ {\sqrt 2 } \right] + \left[ {\sqrt 3 } \right] + ... + \left[ {\sqrt {n^2 - 1} } \right] = \frac{{n(n - 1)(4p + 1)}}{6}<br />
\<br />
    pls solve it with explanation.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    Either I don't know what "p" is or that just makes no sense. That's an awful lot of unrelated irrational numbers magically adding up to a Rational.

    There must be a typo or simply proving it incorrect is easy enough.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    What's that p doing there .. I'm guessing it should be an 'n'. In any case, this sounds like a proof by induction.

    Let P_{n} be the statement that \left[\sqrt{1}\right] + \left[\sqrt{2}\right] + ... + \left[\sqrt{n^2 - 1}\right] = \frac{n(n-1)(4n+1)}{6}.

    Base case: P(1): 0 = 0. True.

    Inductive step: Assume P_{k} to be true. [tex] We must show that P_{k+1} is also true, i.e.
    \left[\sqrt{1}\right] + \left[\sqrt{2}\right] + ... + \left[\sqrt{(k+1)^2 - 1}\right] = \frac{(k+1)((k+1)-1)(4(k+1)+1)}{6}  {\color{white}.}\: \: = \: \: \frac{(k+1)(k)(4k+5)}{6}

    Looking at the left hand side (LHS)
    \text{LHS} = \left[\sqrt{1}\right] + \left[\sqrt{2}\right] + ... + \left[\sqrt{k^{2} + 2k}\right]
    \text{LHS} = {\color{red}\left[\sqrt{1}\right] + \left[\sqrt{2}\right] + ... + \left[\sqrt{k^2 - 1}\right] +} \: \left[\sqrt{k^2}\right] + \left[\sqrt{k^2 + 1}\right] + ... + \left[\sqrt{k^2 + 2k - 1}\right] + \left[\sqrt{k^2 + 2k}\right]

    We see that \left[\sqrt{k^2} \right] = k and the next integer up would be \left[\sqrt{(k+1)^2}\right] = \left[\sqrt{k^2 + 2k + 1}\right] = k + 1. So, since our last term is k^2 + 2k (one less), then we see that : \left[\sqrt{k^2}\right] = \left[\sqrt{k^2 + 1}\right] = ... = \left[\sqrt{k^2 + 2k}\right] = k

    Now, rewriting the equation and noticing that what is in red is P_{k}

    \text{LHS} = {\color{red}\frac{k(k-1)(4k+1)}{6}} + \underbrace{k + k + ... + k}_{\text{(2k + 1) times}}
    \text{LHS} = {\color{red}\frac{k(k-1)(4k+1)}{6}} + (2k+1)k

    Now hopefully if you simplify it, you'll get what you want.

    Phew!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    Ah! The integer function strikes again. I wish that notation were clear or standard.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Aug 2008
    Posts
    21
    Excuse me it's "n" not "p".
    I have one demnostration that not need induction but im not under stand it.i will post the solution and i expect you to explain to me:
    the demonstration become:
    \<br />
\sum\limits_{k = 1}^{n^2  - 1} {\left[ {\sqrt k } \right]}  = \sum\limits_{k = 1}^{n - 1} {k\left[ {(k + 1)^2  - k^2 } \right]}  = \sum\limits_{k = 1}^{n - 1} {2k^2  + k = ....} <br />
\<br />
    Im not understand why???can you explain to me???
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,801
    Thanks
    1691
    Awards
    1
    Quote Originally Posted by o_O View Post
    We see that \left[\sqrt{k^2} \right] = k and the next integer up would be \left[\sqrt{(k+1)^2}\right] = \left[\sqrt{k^2 + 2k + 1}\right] = k + 1. So, since our last term is k^2 + 2k (one less), then we see that :  \underbrace{\left[\sqrt{k^2}\right] = \left[\sqrt{k^2 + 1}\right] = ... = \left[\sqrt{k^2 + 2k}\right] = k}
    Quote Originally Posted by Ortega View Post
    \sum\limits_{k = 1}^{n^2  - 1} {\left[ {\sqrt k } \right]}  = \sum\limits_{k = 1}^{n - 1} {k\left[ {(k + 1)^2  - k^2 } \right]}  = \sum\limits_{k = 1}^{n - 1} {2k^2  + k = ....}
    Im not understand why???can you explain to me???
    Did you read and try to understand o_O reply quoted above.
    That is the explanation!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Aug 2008
    Posts
    21
    Quote Originally Posted by Plato View Post
    That is the explanation!
    yes i undetsand what he wrote.it's more simple than :
    .....(what i wrote)
    But in my demonstration(in the book)they are not precise anything of that
    they write directly the solution .
    i thought that :
    \<br />
\sum {} <br />
\<br />
    have a special proprities,for this reason i ask you again:
    have ...a special proprities???
    for the exemple:
    how we can sum that:
    \<br />
\sum\limits_{k = 1}^{n^2 - 1} {\sqrt k } + \sum\limits_{k = 1}^{n^2 - 1} k <br />
\<br />
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum