Results 1 to 3 of 3

Thread: Square route question

  1. #1
    Senior Member Mukilab's Avatar
    Joined
    Nov 2009
    Posts
    468

    Square route question

    Find the pattern or general rule between $\displaystyle \sqrt{2*3*4*5+1}$ and $\displaystyle \sqrt{0*1*2*3+1}$ and $\displaystyle \sqrt{3*4*5*6+1}$

    I can't see any pattern between them, let alone working out a general rule. Can somebody give me a hint, not the full answer, just a flavour please
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by Mukilab View Post
    Find the pattern or general rule between $\displaystyle \sqrt{2*3*4*5+1}$ and $\displaystyle \sqrt{0*1*2*3+1}$ and $\displaystyle \sqrt{3*4*5*6+1}$

    I can't see any pattern between them, let alone working out a general rule. Can somebody give me a hint, not the full answer, just a flavour please
    List them out...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, Mukilab!

    Find the pattern or general rule for: .$\displaystyle \begin{Bmatrix} \sqrt{0\!\cdot\!1\!\cdot\!2\!\cdot\!3+1} \\ \sqrt{1\!\cdot\!2\!\cdot\!3\!\cdot\!4+1} \\ \sqrt{2\!\cdot\!3\!\cdot\!4\!\cdot\!5+1} \\ \sqrt{3\!\cdot\!4\!\cdot\!5\!\cdot\!6+1} \end{Bmatrix}$

    We have:

    . . $\displaystyle \begin{array}{ccccccc}\sqrt{0\!\cdot\!1\!\cdot\!2\ !\cdot\!3 + 1} &=^& \sqrt{1} &=& 1 \\
    \sqrt{1\!\cdot\!2\!\cdot\!3\!\cdot\!4+1} &=& \sqrt{25} &=& 5 \\
    \sqrt{2\!\cdot\!3\!\cdot\!4\!\cdot\!5+1} &=& \sqrt{121} &=& 11 \\
    \sqrt{3\!\cdot\!4\!\cdot\!5\!\cdot\!6+1} &=& \sqrt{361} &=& 19 \\
    \sqrt{4\!\cdot\!5\!\cdot\!6\!\cdot\!67 + 1} &=& \sqrt{841} &=& 29 \\ \vdots && \vdots \end{array}$


    The general term is:

    . . $\displaystyle a_n \;=\;\sqrt{n(n+1)(n+2)(n+3)+1}\;\text{ for } n = 0,1,2,\hdots $

    . . . . .$\displaystyle =\; \sqrt{n^4 + 6n^3 + 11n^2 + 6n + 1}$

    . . . . .$\displaystyle =\;\sqrt{(n^2 + 3n + 1)^2}$

    . . . . .$\displaystyle =\; n^2+3n+1$

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Quickest Route
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Nov 30th 2010, 05:18 PM
  2. Route cubed
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Apr 21st 2010, 01:56 AM
  3. root mean square question..
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jan 24th 2010, 01:34 AM
  4. minimum spanning tree/shortest route problem
    Posted in the Math Topics Forum
    Replies: 0
    Last Post: Jan 7th 2010, 02:30 AM
  5. Chi Square Problem Question!!
    Posted in the Statistics Forum
    Replies: 0
    Last Post: Apr 15th 2008, 11:52 AM

Search Tags


/mathhelpforum @mathhelpforum