Results 1 to 6 of 6

Math Help - Summation of a geometric sequence

  1. #1
    Newbie
    Joined
    Feb 2010
    Posts
    3

    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:

    S(1) = 1
    S(n) = 5*S(n-1) + 3

    \forall n \geq 2
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    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:

    S(1) = 1
    S(n) = 5*S(n-1) + 3

    \forall n \geq 2
    a_n=\frac{1}{4}\left(7\cdot 5^n-3\right). Care to guess why?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2010
    Posts
    3
    I'd like to take a guess but I honestly don't have a clue on how to do this. At all.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,686
    Thanks
    617
    Herlo!

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


    I found it like this . . .

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

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


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

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

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


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

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

    Divide by 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: . a(n) \;=\;A(1^n) + B(5^n) \;=\;A + (5^n)B


    We know the first two terms of the sequence:

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

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

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


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

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

    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Soroban View Post
    Herlo!

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


    I found it like this . . .

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

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


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

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

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


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

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

    Divide by 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: . a(n) \;=\;A(1^n) + B(5^n) \;=\;A + (5^n)B


    We know the first two terms of the sequence:

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

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

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


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

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

    I believe Soroban meant Drexel \color{red}\bold{28} . That was pretty much what I was going to do, but as usual Soroban applied his uncanny gift for clarity.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Geometric Progression and Summation
    Posted in the Calculus Forum
    Replies: 4
    Last Post: June 10th 2011, 06:44 AM
  2. Summation sequence
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 18th 2010, 11:37 AM
  3. Sequence Summation
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: January 29th 2010, 04:42 PM
  4. Replies: 2
    Last Post: January 23rd 2007, 08:47 AM
  5. Replies: 12
    Last Post: November 15th 2006, 12:51 PM

Search Tags


/mathhelpforum @mathhelpforum