# Thread: Summation of a geometric sequence

1. ## 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:

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

$\displaystyle \forall n \geq 2$

2. 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:

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

$\displaystyle \forall n \geq 2$
$\displaystyle a_n=\frac{1}{4}\left(7\cdot 5^n-3\right)$. Care to guess why?

3. I'd like to take a guess but I honestly don't have a clue on how to do this. At all.

4. 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?

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

6. Originally Posted by Soroban
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.