The constant difference between each term is not $b$, and the first term is not $1$ (but rather $a+b$).
For clarity, I would recommend doing the following. Split the sum into two parts:
$$\sum_{i=1}^n ai + \sum_{i=1}^n b$$
This is equivalent to:
$$[a + 2a + 3a + \dots + (n-1)a + na] + [b + b + b + \dots + b +b]$$
The first sum is an arithmetic progression (A.P) where the difference between each term is $a$, and we have $n$ terms. You seem to know how to do this? The second sum is just the summing of $n$ terms of $b$, which is $n \cdot b$.
The characteristic equation $x^2-\frac{2}{3}x-\frac{1}{3}=0$ has two roots: 1 and $-\frac{1}{3}$.
$$
R_{n+1} + \frac 13 R_n = R_n + \frac 13 R_{n-1} = \cdots = R_1 + \frac{1}{3} R_0 = \frac{17}{30} \tag 1
$$
$$
R_{n+1} - R_n = -\frac{1}{3} (R_n-R_{n-1}) =\cdots = (-\frac 13)^n (R_1-R_0) = -\frac{(-1)^n}{10 \cdot 3^n}\tag 2
$$
(1)-(2)
$$
R_n=\frac{3}{4}\left(\frac{17}{30}+\frac{(-1)^n}{10\cdot 3^n}\right).
$$
Please check my post Show that for every positive integer $ f_n=\frac{\left ( \frac{1+\sqrt5}{2} \right )^n-\left ( \frac{1-\sqrt5}{2} \right )^n}{\sqrt5}$
The above method does not come from nowhere. For a homogeneous second order linear difference equation $$a_{n+2}-b a_{n+1} + c a_n=0 \tag 3$$
its characteristic equation is $f(x)=x^2-bx+c=0$. (3) can be written in the following form:
$$
f(\mathbb{E})a_n=(\mathbb{E}^2-b\mathbb{E} + c)a_n=0 \tag 4
$$
where $\mathbb{E}$ is the forward shift operator such that $\mathbb{E} a_n=a_{n+1}, \mathbb{E}^2 a_n=a_{n+2}$.
Lemma: The solution to $(\mathbb{E}-\lambda)a_n=0$ is $a_n=\lambda^n a_0.$
This is trivial because $(\mathbb{E}-\lambda)a_n=0 \Rightarrow a_{n+1} = \lambda a_n$ which means $a_n$ is a geometric sequence.
Suppose $f(x)=0$ has two distinct roots $r, s$. Then $b=r+s, c=rs$ via Vieta's formulas. Then (3) and (4) become the following, respectively
$$a_{n+2}-(r+s)a_{n+1}+rs a_n=0,\tag 5$$
$$f(\mathbb{E})a_n=(\mathbb{E}-r)(\mathbb{E} - s)a_n=0. \tag 6$$
In the following table you will see that the "shortcut" method is simply a factorization of the characteristic equation in terms of $\mathbb{E}$.
$$
\begin{array}{lcl}
a_{n+1} - s a_n = r(a_n-sa_{n-1}) & | & (\mathbb{E}-r) (\mathbb{E}-s)a_n=0\\
\Rightarrow \color{red}{a_{n+1} - s a_n = r^n (a_1-s a_0)} & | & \Rightarrow \color{red}{(\mathbb{E}-s)a_n = r^n (\mathbb{E}-s)a_0} \text{ via Lemma} \\
a_{n+1} - r a_n = s(a_n-ra_{n-1}) & | & (\mathbb{E}-s) (\mathbb{E}-r)a_n=0\\
\Rightarrow \color{blue}{a_{n+1} - r a_n = s^n (a_1-ra_0)} & | & \Rightarrow \color{blue}{(\mathbb{E}-r)a_n = s^n (\mathbb{E}-r)a_0} \text{ via Lemma} \\
\end{array}
$$
Subtracting the blue equation from the red, you get
$$
a_n=\frac{a_1-s a_0}{r-s} r^n - \frac{a_1-r a_0}{r-s} s^n
$$
Solution with duplicate roots
There is also a shortcut. If $f(x)=(x-\lambda)^2$, or
$$a_{n+1} - 2\lambda a_n + \lambda^2 a_{n-1}=0.$$
If $\lambda=0$ it's trivial: $a_n=0$. Otherwise $\lambda \neq 0$, then
$$\frac{a_{n+1}}{\lambda^{n+1}} - 2 \frac{a_n}{\lambda^n} + \frac{a_{n-1}}{\lambda^{n-1}}=0 \Rightarrow \frac{a_{n+1}}{\lambda^{n+1}} - \frac{a_n}{\lambda^n} = \frac{a_n}{\lambda^n} - \frac{a_{n-1}}{\lambda^{n-1}} = \cdots =\frac{a_1}{\lambda} - a_0$$
So $\frac{a_n}{\lambda^n}$ is an arithmetic sequence,
$$
\frac{a_n}{\lambda^n}=\frac{a_0}{\lambda^0}+n\left( \frac{a_1}{\lambda} - a_0\right) \Rightarrow a_n = \lambda^n(na_1/\lambda-(n-1) a_0).
$$
Other examples
Some non-homogeneous linear difference equations can be converted to homogeneous higher order ones.
Example 1: Recurring Sequence with Exponent
Example 2: $a_n=3a_{n-1}+1$.
For this one, although we can convert to $a_{n+1} - 3a_{n} = a_n-3 a_{n-1}$, it's easier to do the following: $a_n + \frac 12 =3a_{n-1}+\frac{3}{2} = 3 (a_{n-1} + \frac 12) \Rightarrow a_n+\frac 12 = 3^n (a_0+\frac{1}{2}).$
Example 3 (a higher order example illustrating the use of forward shift operator): Is it possible to solve this recurrence equation?
Best Answer
You could use generating functions to solve the recurrence relation. We use GFs as formal power series i.e. purely algebraically without considering convergence or limits. A formidable starter for these techniques is H.Wilf's Generatingfunctionology.
Let's consider the generating function
and use the Ansatz:
Comment:
In (2) we made an index shift $n\rightarrow n-1$
In (3) you may observe that the right sum is the derivative of the geometrical series $\frac{1}{1-z}$
In (4) we apply the formal differential operator $D$
Now we calculate $F(z)$ and find a series representation.
Observe, that we applied partial fraction decomposition in (5).