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Thread: Sequence

  1. #1
    Senior Member TheAbstractionist's Avatar
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    Sequence

    Hi all.

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

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

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

    and so on.

    Prove that

    $\displaystyle a_n\ =\ \frac{2\sqrt{6^{n-2}}+\sqrt{6^{n-1}}+(-1)^n\left(2\sqrt{6^{n-2}}-\sqrt{6^{n-1}}\right)}2$
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  2. #2
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    isn't $\displaystyle a_n=2^{\lfloor \frac{n}{2} \rfloor} 3^{\lfloor \frac{n-1}{2} \rfloor}$ a better looking formula for your sequence?
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  3. #3
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    isn't $\displaystyle a_n=2^{\lfloor \frac{n}{2} \rfloor} 3^{\lfloor \frac{n-1}{2} \rfloor}$ a better looking formula for your sequence?
    Hi NonCommAlg.

    No, it isnít.
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  4. #4
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    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?
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    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
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  6. #6
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    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.
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  7. #7
    Moo
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    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 ?
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  8. #8
    MHF Contributor red_dog's Avatar
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    We have $\displaystyle a_{n+2}-6a_n=0, \ \forall n\geq 3$

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

    We try to find the general term $\displaystyle a_n$ as

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

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

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

    Then we replace $\displaystyle \alpha, \ \beta, \ x_1, \ x_2$ and we get the formula for $\displaystyle a_n$
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  9. #9
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by Jhevon View Post
    we want to make the students sweat a little bit
    Hi Jhevon.

    Yes, thatís the point!
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  10. #10
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by red_dog View Post
    We have $\displaystyle a_{n+2}-6a_n=0, \ \forall n\geq 3$

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

    We try to find the general term $\displaystyle a_n$ as

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

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

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

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

    Thanks for your solution. Here is mine.

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

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

    Now

    $\displaystyle a_n\ =\ \underbrace{\frac{1-(-1)^n}2a_n}_{\mbox{odd terms}}\,+\,
    \underbrace{\frac{1+(-1)^n}2a_n}_{\mbox{even terms}}$


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

    Substituting $\displaystyle b_{\frac{n+1}2}=6^{\frac{n+1}2-1}=\sqrt{6^{n-1}}$ and $\displaystyle c_{\frac n2}=2\cdot6^{\frac n2-1}=2\sqrt{6^{n-2}}$ will give the required formula for $\displaystyle a_n$.
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  11. #11
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    I get this part:




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



    Thanks for your help!!
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  12. #12
    Senior Member TheAbstractionist's Avatar
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    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 $\displaystyle n$ is odd, then $\displaystyle \frac{1+(-1)^n}2a_n$ disappears, leaving the other half. If $\displaystyle n$ 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

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

    Here

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

    Hence $\displaystyle a_n=\frac{1-(-1)^n}2b_{\frac{n-1}2}+\frac{1+(-1)^n}2c_{\frac n2}=\frac{\sqrt n+n+(-1)^n(\sqrt n-n)}2$
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