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Thread: Summation of a geometric sequence

  1. #1
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    Summation of a geometric sequence

    I need to find the closed-form solution to the following problem. I'm not really sure how to do the summation of a geometric sequence. Anyways, here it is:

    $\displaystyle S(1) = 1$
    $\displaystyle S(n) = 5*S(n-1) + 3$

    $\displaystyle \forall n \geq 2$
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Neodymium View Post
    I need to find the closed-form solution to the following problem. I'm not really sure how to do the summation of a geometric sequence. Anyways, here it is:

    $\displaystyle S(1) = 1$
    $\displaystyle S(n) = 5*S(n-1) + 3$

    $\displaystyle \forall n \geq 2$
    $\displaystyle a_n=\frac{1}{4}\left(7\cdot 5^n-3\right)$. Care to guess why?
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  3. #3
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    I'd like to take a guess but I honestly don't have a clue on how to do this. At all.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Neodymium View Post
    I'd like to take a guess but I honestly don't have a clue on how to do this. At all.
    What is your background in recurrence relations?
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  5. #5
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    Herlo!

    I believe Drexel128 meant: .$\displaystyle \frac{1}{4}\left(7\cdot5^{{\color{red}n-1}}-3\right)$


    I found it like this . . .

    From: .$\displaystyle a_1 \:=\:1,\;\;a_n \:=\:5\!\cdot\!a_{n-1} + 3$

    . . we have: .$\displaystyle \begin{Bmatrix} a_1 &=& 1 \\ a_2 &=& 8 \\ \vdots && \vdots \end{Bmatrix}$


    $\displaystyle \begin{array}{cccccc}\text{We have:} & a_n &=& 5\cdot a_{n-1} + 3 & [1] \\
    \text{Next term: } & a_{n+1} &=& 5\cdot a_n + 3 & [2] \end{array}$

    Subtract [2] - [1]:] .$\displaystyle a_{n+1} - a_n \;=\;5\cdot a_n - 5a_{n-1} $

    . . . and we have: . $\displaystyle a_{n+1} - 6a_n + 5a_{n-1} \:=\:0\;\;[3]$


    Let: .$\displaystyle X^n \,=\,a_n$

    Then [3] becomes: .$\displaystyle X^{n+1} - 6X^n + 5X^{n-1} \;=\;0$

    Divide by $\displaystyle X^{n-1}\!:\;\;X^2 - 6X + 5 \;=\;0 \quad\Rightarrow\quad (X - 1)(X - 5) \:=\:0 \quad\Rightarrow\quad X \:=\:1,\:5$


    The generating function has the form: .$\displaystyle a(n) \;=\;A(1^n) + B(5^n) \;=\;A + (5^n)B $


    We know the first two terms of the sequence:

    . . $\displaystyle \begin{array}{ccccccc}a(1) = 1\!: & A + 5B &=& 1 & [1] \\
    a(2) = 8\!: & A + 25B &=& 8 & [2] \end{array}$

    Sugbtract [2] - [1]: .$\displaystyle 20B \:=\:7 \quad\Rightarrow\quad B \:=\:\frac{7}{20}$

    Sustitute into [1]: .$\displaystyle A + 5\left(\tfrac{7}{20}\right) \:=\:1 \quad\Rightarrow\quad A \:=\;-\frac{3}{4}$


    Hence, the function is: .$\displaystyle a(n) \;\;=\;\;\frac{7}{20}\cdot 5^n - \frac{3}{4} \;\;=\;\;\frac{7}{4}\cdot5^{n-1} - \frac{3}{4}$

    . . Therefore: .$\displaystyle a(n) \;\;=\;\;\frac{1}{4}\left(7\cdot5^{n-1} - 3\right)$

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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Soroban View Post
    Herlo!

    I believe Drexel128 meant: .$\displaystyle \frac{1}{4}\left(7\cdot5^{{\color{red}n-1}}-3\right)$


    I found it like this . . .

    From: .$\displaystyle a_1 \:=\:1,\;\;a_n \:=\:5\!\cdot\!a_{n-1} + 3$

    . . we have: .$\displaystyle \begin{Bmatrix} a_1 &=& 1 \\ a_2 &=& 8 \\ \vdots && \vdots \end{Bmatrix}$


    $\displaystyle \begin{array}{cccccc}\text{We have:} & a_n &=& 5\cdot a_{n-1} + 3 & [1] \\
    \text{Next term: } & a_{n+1} &=& 5\cdot a_n + 3 & [2] \end{array}$

    Subtract [2] - [1]:] .$\displaystyle a_{n+1} - a_n \;=\;5\cdot a_n - 5a_{n-1} $

    . . . and we have: . $\displaystyle a_{n+1} - 6a_n + 5a_{n-1} \:=\:0\;\;[3]$


    Let: .$\displaystyle X^n \,=\,a_n$

    Then [3] becomes: .$\displaystyle X^{n+1} - 6X^n + 5X^{n-1} \;=\;0$

    Divide by $\displaystyle X^{n-1}\!:\;\;X^2 - 6X + 5 \;=\;0 \quad\Rightarrow\quad (X - 1)(X - 5) \:=\:0 \quad\Rightarrow\quad X \:=\:1,\:5$


    The generating function has the form: .$\displaystyle a(n) \;=\;A(1^n) + B(5^n) \;=\;A + (5^n)B $


    We know the first two terms of the sequence:

    . . $\displaystyle \begin{array}{ccccccc}a(1) = 1\!: & A + 5B &=& 1 & [1] \\
    a(2) = 8\!: & A + 25B &=& 8 & [2] \end{array}$

    Sugbtract [2] - [1]: .$\displaystyle 20B \:=\:7 \quad\Rightarrow\quad B \:=\:\frac{7}{20}$

    Sustitute into [1]: .$\displaystyle A + 5\left(\tfrac{7}{20}\right) \:=\:1 \quad\Rightarrow\quad A \:=\;-\frac{3}{4}$


    Hence, the function is: .$\displaystyle a(n) \;\;=\;\;\frac{7}{20}\cdot 5^n - \frac{3}{4} \;\;=\;\;\frac{7}{4}\cdot5^{n-1} - \frac{3}{4}$

    . . Therefore: .$\displaystyle a(n) \;\;=\;\;\frac{1}{4}\left(7\cdot5^{n-1} - 3\right)$

    I believe Soroban meant Drexel$\displaystyle \color{red}\bold{28}$ . That was pretty much what I was going to do, but as usual Soroban applied his uncanny gift for clarity.
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