Results 1 to 4 of 4

Math Help - hexagon identity

  1. #1
    Member
    Joined
    May 2008
    Posts
    94

    Post hexagon identity

    Can anyone prove the below given identity?
    I am giving it a try I also want someone else to verify!

    Suppose that k and n are integers with 1<k<=n. Prove the hexagon identity
    which relates terms in Pascal's triangle that form a hexagon.



    <br />
\left(\begin{array}{cc}n-1\\k-1\end{array}\right)  \left(\begin{array}{cc}n\\k+1\end{array}\right)<br />
\left(\begin{array}{cc}n+1\\k\end{array}\right) = <br />
\left(\begin{array}{cc}n-1\\k\end{array}\right)<br />
 \left(\begin{array}{cc}n\\k-1\end{array}\right)<br />
\left(\begin{array}{cc}n+1\\k+1\end{array}\right)<br />
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,929
    Thanks
    333
    Awards
    1
    Quote Originally Posted by robocop_911 View Post
    Can anyone prove the below given identity?
    I am giving it a try I also want someone else to verify!

    Suppose that k and n are integers with 1<k<=n. Prove the hexagon identity
    which relates terms in Pascal's triangle that form a hexagon.



    <br />
\left(\begin{array}{cc}n-1\\k-1\end{array}\right)  \left(\begin{array}{cc}n\\k+1\end{array}\right)<br />
\left(\begin{array}{cc}n+1\\k\end{array}\right) = <br />
\left(\begin{array}{cc}n-1\\k\end{array}\right)<br />
 \left(\begin{array}{cc}n\\k-1\end{array}\right)<br />
\left(\begin{array}{cc}n+1\\k+1\end{array}\right)<br />
    The easiest way is to just write out the terms for each side:
    Left hand side:
    {{n - 1} \choose {k - 1}} {{n} \choose {k + 1}} {{n + 1} \choose {k}} = \frac{(n - 1)!}{(k - 1)!(n - k)!} \cdot \frac{n!}{(k + 1)!(n - k - 1)!} \cdot \frac{(n + 1)!}{k!(n + 1 - k)!}

    You do the right hand side.

    By the way, check out how I did the coding. It's much easier.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2008
    Posts
    94
    Quote Originally Posted by topsquark View Post
    The easiest way is to just write out the terms for each side:
    Left hand side:
    {{n - 1} \choose {k - 1}} {{n} \choose {k + 1}} {{n + 1} \choose {k}} = \frac{(n - 1)!}{(k - 1)!(n - k)!} \cdot \frac{n!}{(k + 1)!(n - k - 1)!} \cdot \frac{(n + 1)!}{k!(n + 1 - k)!}

    You do the right hand side.
    -Dan

    Can we "simplify" this identity a bit more? Or we just have to convert this identity into "factorial" form and we are done?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,929
    Thanks
    333
    Awards
    1
    Quote Originally Posted by robocop_911 View Post
    Can we "simplify" this identity a bit more? Or we just have to convert this identity into "factorial" form and we are done?
    Why simplify? All you need to do is show that both sides are equal. When you write out the right hand side you will see that you have the same terms in the numerator and denominator.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Hexagon
    Posted in the Geometry Forum
    Replies: 0
    Last Post: October 15th 2010, 05:45 PM
  2. hexAGON
    Posted in the Geometry Forum
    Replies: 5
    Last Post: October 16th 2009, 09:00 AM
  3. hexagon
    Posted in the Algebra Forum
    Replies: 0
    Last Post: November 11th 2007, 08:36 AM
  4. Hexagon
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 30th 2006, 02:46 AM
  5. **Hexagon Help**
    Posted in the Geometry Forum
    Replies: 11
    Last Post: May 2nd 2006, 09:26 AM

Search Tags


/mathhelpforum @mathhelpforum