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Math Help - really confused with sequences, need some help

  1. #1
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    really confused with sequences, need some help

    hi all,

    we had a "substitute professor" for our discrete class the other day, and to say the least she did a horrible job. I really do not understand how you solve this question at all. Any pointers will be very greatly appreciated. I don't even know where to start.

    really confused with sequences, need some help-untitled.png
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  2. #2
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    Re: really confused with sequences, need some help

    You find it one number at a time. For (a), for example, find $a_1$ by plugging in $1$ everywhere you see an $n$. So, $a_1 = 7a_{1-1}-5 = 7a_0-5$. You are given that $a_0 = 1$, so $a_1 = 7(1)-5 = 2$. Then plug in $2$ for $n$: $a_2 = 7a_{2-1} - 5 = 7a_1-5 = 7(2) - 5 = 9$. Can you figure out $a_3, a_4, a_5$ from there?

    Use the same process for (b) through (d).
    Last edited by SlipEternal; March 25th 2014 at 06:01 PM.
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  3. #3
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    Re: really confused with sequences, need some help

    Quote Originally Posted by SlipEternal View Post
    You find it one number at a time. For (a), for example, find $a_1$ by plugging in $1$ everywhere you see an $n$. So, $a_1 = 7a_{1-1}-5 = 7a_0-5$. You are given that $a_0 = 1$, so $a_1 = 7(1)-5 = 2$. Then plug in $2$ for $n$: $a_2 = 7a_{2-1} - 5 = 7a_1-5 = 7(2) - 5 = 9$. Can you figure out $a_3, a_4, a_5$ from there?

    Use the same process for (b) through (d).

    I believe I understand it, so, for a3 = 7a3-1 - 5 = 7a2-5 = 7(9) - 5 = 58

    would that be correct?
    Last edited by michaelgg13; March 25th 2014 at 06:23 PM.
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  4. #4
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    Re: really confused with sequences, need some help

    Quote Originally Posted by michaelgg13 View Post
    I believe I understand it, so, for a3 = 7a3-1 - 5 = 7a2-5 = 7(9) - 5 = 58$

    would that be correct?
    Yes, that is correct.
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  5. #5
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    Re: really confused with sequences, need some help

    Thank you, and I am slightly confused by C still.

    a1=a1-1-3a1-2=a0-3a-1

    but -1 isn't defined and it also says for n>1
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  6. #6
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    Re: really confused with sequences, need some help

    You are given $a_1$ already. Start calculating $a_2$.
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  7. #7
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    Re: really confused with sequences, need some help

    this is a little off topic, but would you happen to be able to explain how you compute this summation?

    really confused with sequences, need some help-untitled.png
    Last edited by michaelgg13; March 26th 2014 at 07:17 AM.
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  8. #8
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    Re: really confused with sequences, need some help

    That is known as a geometric sum. In general, $\dfrac{a^{n+1}-1}{a-1} = \sum_{k=0}^n a^k$

    If you want to see the proof of this, do a google search on proof of the geometric sum. In your case, you have a sum that is similar to the right-hand side of the equation. But, instead of $a^k$ inside the summation, you have $\dfrac{8^k}{5^{k+2}}$. You need to play with it to make it look like the formula. Now, $5^{k+2}=5^k\cdot 5^2$, so you can rewrite the summation:

    $\displaystyle \begin{align*} \sum_{k=0}^{40} \dfrac{ 8^k }{ 5^{k+2} } & = \sum_{k=0}^{40} \dfrac{ 1 }{ 5^2 } \cdot \dfrac{ 8^k }{ 5^k } \\ & = \dfrac{ 1 }{ 5^2 } \sum_{k=0}^{40} \left( \dfrac{ 8 }{ 5 } \right)^k \end{align*}$

    Then let $a=\dfrac{8}{5}$ and plug it into the geometric sum formula above.
    Last edited by SlipEternal; March 26th 2014 at 07:24 AM.
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  9. #9
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    Re: really confused with sequences, need some help

    Quote Originally Posted by michaelgg13 View Post
    this is a little off topic, but would you happen to be able to explain how you compute this summation?

    Click image for larger version. 

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    $\displaystyle{\sum_{k=0}^{40}}\dfrac{8^k}{5^{k+2} }=$

    $\displaystyle{\sum_{k=0}^{40}}\dfrac{8^k}{5^k \cdot 25}=$

    $\dfrac{1}{25}\displaystyle{\sum_{k=0}^{40}}\left( \dfrac{8}{5}\right)^k=$

    $\dfrac{1}{25}\dfrac{1-\left(\frac{8}{5}\right)^{41}}{1-\frac{8}{5}}=$

    $\dfrac{1}{25}\dfrac{\left(\frac{8}{5}\right)^{41}-1}{\frac{3}{5}}=$

    $\dfrac{\left(\frac{8}{5}\right)^{41}-1}{15}$
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  10. #10
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    Re: really confused with sequences, need some help

    Hello, michaelgg13!

    Perhaps you are confused by the notation?



    \text{Find }a_5\text{ given: }\:a_n \:=\:7a_{n-2} - 5,\;a_0 = 1

    We have a sequence of numbers: . a_0,\,a_1,\,a_2,\,a_3,\,\hdots

    It begins with a_0 = 1.

    The succesive terms are given by: . a_n \:=\:7a_{n-1}-5

    This means: \text{the }n^{th}\text{ term is "7 times the previous term, minus 5".}


    \begin{array}{cccccccc}\text{So that:} & a_0 &=& 1 \\ & a_1 &=& 7(1) - 5 &=& 2 \\ & a_2 &=& 7(2)-5 &=& 9 \\ & a_3 &=& 7(9)-5 &=& 58 \\ & a_4 &=& 7(58)-5 &=& 401 \\ & a_5 &=& 7(401)-5 &=& 2802 \end{array}


    Get the idea?


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    \text{FYI: }\,\text{The }n^{th}\text{ term is: }\:a_n \;=\;\tfrac{1}{6}(7^n+5)
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