Results 1 to 12 of 12

Math Help - Sequence

  1. #1
    Senior Member TheAbstractionist's Avatar
    Joined
    Apr 2009
    Posts
    328
    Thanks
    1

    Sequence

    Hi all.

    This is a simple challenge problem I made up myself. Consider the sequence \left(a_n\right)=(1,2,6,12,36,72,\ldots).

    The first term is 1, and successive terms are formed by multiplying the previous term by 2 and 3 alternately, starting with 2. Thus

    a_1=1
    a_2=1\cdot2=2
    a_3=2\cdot3=6
    a_4=6\cdot2=12
    a_5=12\cdot3=36
    a_6=36\cdot2=72

    and so on.

    Prove that

    a_n\ =\ \frac{2\sqrt{6^{n-2}}+\sqrt{6^{n-1}}+(-1)^n\left(2\sqrt{6^{n-2}}-\sqrt{6^{n-1}}\right)}2
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    isn't a_n=2^{\lfloor \frac{n}{2} \rfloor} 3^{\lfloor \frac{n-1}{2} \rfloor} a better looking formula for your sequence?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member TheAbstractionist's Avatar
    Joined
    Apr 2009
    Posts
    328
    Thanks
    1
    Quote Originally Posted by NonCommAlg View Post
    isn't a_n=2^{\lfloor \frac{n}{2} \rfloor} 3^{\lfloor \frac{n-1}{2} \rfloor} a better looking formula for your sequence?
    Hi NonCommAlg.

    No, it isnít.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by TheAbstractionist View Post

    No, it isnít.
    why not? haha

    i have a question for computer programmers: if you had to use one of these two formulas, which one would you choose and why?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by NonCommAlg View Post
    why not? haha

    i have a question for computer programmers: if you had to use one of these two formulas, which one would you choose and why?
    well, i'm not a programmer, but your formula would be used i bet. but seeing that this is a challenge problem, your formula seems too intuitively obvious (and probably easy to prove. induction would work, right?). we want to make the students sweat a little bit
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Jhevon View Post

    but seeing that this is a challenge problem, your formula seems too intuitively obvious (and probably easy to prove. induction would work, right?). we want to make the students sweat a little bit
    yeah, it's just a simple induction.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Jhevon View Post
    (and probably easy to prove. induction would work, right?).
    Spoiler:
    and separating the cases n odd, n even... it should be obvious ?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    We have a_{n+2}-6a_n=0, \ \forall n\geq 3

    Consider the characteristic equation x^2-6=0 with solutions x_1=-\sqrt{6}, \ x_2=\sqrt{6}

    We try to find the general term a_n as

    a_n=\alpha x_1^n+\beta x_2^n

    where \alpha, \ \beta will be determined knowing the first two terms of the sequence.

    Replacing n=1 and n=2 we get \alpha=\frac{2-\sqrt{6}}{12}, \ \beta=\frac{2+\sqrt{6}}{6}

    Then we replace \alpha, \ \beta, \ x_1, \ x_2 and we get the formula for a_n
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member TheAbstractionist's Avatar
    Joined
    Apr 2009
    Posts
    328
    Thanks
    1
    Quote Originally Posted by Jhevon View Post
    we want to make the students sweat a little bit
    Hi Jhevon.

    Yes, thatís the point!
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member TheAbstractionist's Avatar
    Joined
    Apr 2009
    Posts
    328
    Thanks
    1
    Quote Originally Posted by red_dog View Post
    We have a_{n+2}-6a_n=0, \ \forall n\geq 3

    Consider the characteristic equation x^2-6=0 with solutions x_1=-\sqrt{6}, \ x_2=\sqrt{6}

    We try to find the general term a_n as

    a_n=\alpha x_1^n+\beta x_2^n

    where \alpha, \ \beta will be determined knowing the first two terms of the sequence.

    Replacing n=1 and n=2 we get \alpha=\frac{2-\sqrt{6}}{12}, \ \beta=\frac{2+\sqrt{6}}{6}

    Then we replace \alpha, \ \beta, \ x_1, \ x_2 and we get the formula for a_n
    Hi red_dog.

    Thanks for your solution. Here is mine.

    Let \left(b_n\right) and \left(c_n\right) be the odd and even subsequences respectively of \left(a_n\right). Hence \left(b_n\right) and \left(c_n\right) are geometric progressions with common ratio 6:

    b_n\ =\ a_{2n-1}\ =\ 6^{n-1}
    c_n\ =\ a_{2n}\ =\ 2\cdot6^{n-1}

    Now

    a_n\ =\ \underbrace{\frac{1-(-1)^n}2a_n}_{\mbox{odd terms}}\,+\,<br />
\underbrace{\frac{1+(-1)^n}2a_n}_{\mbox{even terms}}


    ___ =\ \frac{1-(-1)^n}2b_{\frac{n+1}2}\,+\,\frac{1+(-1)^n}2c_{\frac n2}

    Substituting b_{\frac{n+1}2}=6^{\frac{n+1}2-1}=\sqrt{6^{n-1}} and c_{\frac n2}=2\cdot6^{\frac n2-1}=2\sqrt{6^{n-2}} will give the required formula for a_n.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    May 2009
    Posts
    1
    I get this part:




    However, I don't understand where you got this equation =P



    Thanks for your help!!
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Senior Member TheAbstractionist's Avatar
    Joined
    Apr 2009
    Posts
    328
    Thanks
    1
    Quote Originally Posted by weirdswimguy View Post
    I get this part:




    However, I don't understand where you got this equation =P



    Thanks for your help!!
    Hi weirdswimguy.

    If n is odd, then \frac{1+(-1)^n}2a_n disappears, leaving the other half. If n is even, the other half disappears, leaving this one. Neat, yeah?

    This method is very handy for writing down a formula for sequences whose properties alternate between odd and even terms. For example, suppose you have

    (a_n)\ =\ (1,\sqrt2,3,\sqrt4,5,\sqrt6,7,\ldots)

    Here

    b_n=2n-1
    c_n=\sqrt{2n}

    Hence a_n=\frac{1-(-1)^n}2b_{\frac{n-1}2}+\frac{1+(-1)^n}2c_{\frac n2}=\frac{\sqrt n+n+(-1)^n(\sqrt n-n)}2
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 24th 2010, 02:10 AM
  2. Replies: 0
    Last Post: July 4th 2010, 12:05 PM
  3. Replies: 2
    Last Post: March 1st 2010, 11:57 AM
  4. sequence membership and sequence builder operators
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: June 4th 2009, 03:16 AM
  5. Replies: 12
    Last Post: November 15th 2006, 12:51 PM

Search Tags


/mathhelpforum @mathhelpforum