# General term of recurrence relation

Find the general term of the following recurrence relation:

$$a_{1} = 2$$

$$a_{n+1} = \frac{2a_{n} - 1}{3}$$

I've tried to find the first few terms:

$$a_{1} = 2$$ $$a_{2} = \frac{2.2 - 1}{3} = 1$$ $$a_{3} = \frac{2.1 - 1}{3} = \frac{1}{3}$$ $$a_{4} = -\frac{1}{9}$$ $$a_{5} = -\frac{1}{27}$$

but can't see any pattern, especially considering the first 2 terms.

We have $$3a_{n+1}=2a_n-1 \tag{1}$$ and we may try to get rid of the inhomogeneous term $-1$ by a suitable substitution.
For instance, by setting $a_n=b_n-1$, we get $b_1=3$ and $$3b_{n+1}=2b_n \tag{2}$$ from which $b_n = 3\cdot\frac{2^{n-1}}{3^{n-1}}$ readily follows. The general term of the original sequence is so: $$a_n = \color{red}{3\cdot\frac{2^{n-1}}{3^{n-1}}-1}.\tag{3}$$

• Thank you for your time and answer! Could you suggest a source for further reading on this topic? – Ziezi Sep 15 '16 at 12:25
• @Ziezi: these slides nms.lu.lv/wp-content/uploads/2016/04/21-linear-recurrences.pdf should be a good starting point. – Jack D'Aurizio Sep 15 '16 at 12:32
• The slides are great! (just to note) The link at the end is broken. – Ziezi Sep 15 '16 at 18:43

Following your (@Jack D'Aurizio) suggestion, here is what I understood:

The given recurrence relation: $3a_{n} = 2a_{n-1} - 1$, with initial conditions: $a_{1} = 2$ is a $1^{st}$ degree linear non-homogeneous recurrence relation.

Solution:

We are searching for solution1 of the form: $a_{n} = b_{n} + h_{n}$ (1), where $h_{n}$ is the sequence satisfying the associated homogeneous recurrence relation and $b_{n}$ is a solution which is similar to $f(n)$.

1.The associated homogeneous recurrence relation is: $3h_{n} = 2h_{n-1}$ and its characteristic equation is: $3r - 2 = 0$ or $r = \frac{2}{3}$ and the solution is:

$$h_{n} = \alpha_{0} (\frac{2}{3})^{n-1}$$

From the initial conditions: $a_{1} = \alpha_{0} (\frac{2}{3})^{1} = 2$ or $\alpha_{0} = 3$, so the solution of the homogeneous recurrence is:

$$h_{n} = 3 (\frac{2}{3})^{n-1}$$

1. In our case $b_{n} = -1$ and from (1) we have:

$$a_{n} = 3 (\frac{2}{3})^{n-1} - 1$$

1. $a_{n}$ satisfies both the recurrence relation and the initial conditions.

2. $f(n)$ - function depending only on $n$.