I have the following recurrence relations:
$$t_0=\frac{1}{a+b}+\frac{a}{a+b}\frac{1}{c}\\t_n=\frac{1}{a+b}+\frac{a}{a+b}\frac{n+1}{c}+\frac{b}{a+b}\sum_{j=1}^n p^j q^{n-j} t_{n-j}\\with\quad\quad p+q=1.$$ I want to find a closed form expression for $t_n$ (iterative form). I calculated $t_1, t_2, t_3, t_4$ and then try to find the iterative pattern, but I couldn't. The expressions that I obtained are \begin{align}t_1&= \frac{1}{a+b}+\frac{a}{a+b}\frac{2}{c}+\frac{b}{a+b}p~ t_0\\ t_2&=\frac{1}{a+b}\left(1+\frac{pqb}{a+b}\right)+\frac{a}{a+b}\frac{1}{c}\left(3+2\frac{pqb}{a+b}\right)+\frac{b}{a+b}\left(p^2+p^2q\frac{b}{a+b}\right)t_0\\t_3 &=\frac{1}{a+b}\left(1+\frac{p^2qb}{a+b}+\frac{pq^2b}{a+b}+\frac{p^2q^3b^2}{(a+b)^2}\right)+\frac{a}{a+b}\frac{1}{c}\left(4+\frac{b}{a+b}pq^2(3+2\frac{pqb}{a+b})+2p^2q\frac{b}{a+b}\right)+\frac{b}{a+b}\left(\frac{b}{a+b}p^3q^2(1+\frac{qb}{a+b})+p^3q^2\frac{b}{a+b}+p^3\right)t_0. \end{align} From these equations how to obtain the general iterative form?
It seems the following.
Let $n\ge 2$. Then $$\sum_{j=1}^n p^j q^{n-j} t_{n-j}=p\sum_{j=1}^{n-1} p^j q^{n-1-j} t_{n-1-j}+ pq^{n-1} t_{n-1}.$$ So, using the recurrent formula for $t_n$ we obtain
$$t_n-pt_{n-1}=\frac {1-p}{a+b}+\frac{a}{a+b}\frac{n+1}{c}- \frac{pa}{a+b}\frac{n}{c}+ \frac{b}{a+b} pq^{n-1} t_{n-1}.$$
Update. Which yields:
$$t_n=(p+xq^{n-1})t_{n-1}+yn+z,$$
where $x=\frac{bp}{a+b},$ $y=\frac{a(1-p)}{(a+b)c},$ and $z=\frac{c-pc+a}{(a+b)c}.$
Unfortunately, the general formula for $t_n$, which I obtained from it, is essentially as complex as the recurrent one.
$$t_n=t_1\prod_{j=1}^{n-1} (p+xq^j)+\sum_{i=2}^{n} (yi+z)\prod_{j=i}^{n-1} (p+xq^j).$$