How to solve the recurrence relation $t_n=(1+c q^{n-1})p~t_{n-1}+a +nbq$? How to solve $$t_n=(1+c q^{n-1})p~t_{n-1}+a +nbq,\quad n\ge 2.$$ given that  $$t_1=b+(1+c~p)(a~q^{-1}+b),\qquad p+q=1.$$
N.B- Some misprints in the question I corrected. Sorry for the misprint $an+b$ is actually $a+nbq$.

let
$$\dfrac{h_{n-1}}{h_{n}}=(1+cq^{n-1})p$$
so we have 
$$\dfrac{h_{n-2}}{h_{n-1}}=(1+cq^{n-2})p$$
$$\cdots\cdots$$
$$\dfrac{h_{1}}{h_{2}}=(1+cq)p$$
$$\dfrac{h_{1}}{h_{n}}=\prod_{k=1}^{n-1}(1+cq^k)p\Longrightarrow h_{n}=\dfrac{h_{1}}{\prod_{k=1}^{n-1}(1+cq^k)p}$$
then we have
$$t_{n}=\dfrac{h_{n-1}}{h_{n}}t_{n-1}+an+b$$
$$\Longrightarrow h_{n}t_{n}=h_{n-1}t_{n-1}+(an+b)h_{n}$$
so we have
$$h_{n}t_{n}=\sum_{i=2}^{n}\left(h_{i}t_{i}-h_{i-1}t_{i-1}\right)+h_{1}t_{1}=\sum_{i=2}^{n}(ai+b)h_{i}+h_{1}t_{1}$$
so
$$t_{n}=\dfrac{\sum_{i=2}^{n}(ai+b)h_{i}+h_{1}t_{1}}{h_{n}}=\prod_{k=1}^{n-1}(1+cq^k)p\cdot\left(\dfrac{(\sum_{i=2}^{n}(ai+b)h_{i}}{h_{1}}+t_{1}\right)$$
