# Summation of a geometric sequence

• February 17th 2010, 08:23 PM
Neodymium
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:

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

$\forall n \geq 2$
• February 17th 2010, 09:40 PM
Drexel28
Quote:

Originally Posted by Neodymium
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?
• February 18th 2010, 12:05 AM
Neodymium
I'd like to take a guess but I honestly don't have a clue on how to do this. At all.
• February 18th 2010, 11:40 AM
Drexel28
Quote:

Originally Posted by Neodymium
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?
• February 18th 2010, 02:05 PM
Soroban
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: . $\begin{Bmatrix} a_1 &=& 1 \\ a_2 &=& 8 \\ \vdots && \vdots \end{Bmatrix}$

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

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:

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

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

• February 18th 2010, 02:10 PM
Drexel28
Quote:

Originally Posted by Soroban
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: . $\begin{Bmatrix} a_1 &=& 1 \\ a_2 &=& 8 \\ \vdots && \vdots \end{Bmatrix}$

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

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:

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

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.