I've been trying to figure out this recursion problem but I'm getting stuck trying to find the nth-term sequence for the last recursion. I found one but the second i'm so clueless about. I don't know what algebra manipulation I'm missing but could someone shed some light to it? Is it even possible?
$Definition: $
$$w_0 = a$$ $$\forall n \in \mathbb{N}, w_{n+1} = w_n +b $$ $$ z_0 = w_0$$ $$\forall n \in \mathbb{N}, z_{n+1} = z_n + (w_{n+1}) x^n$$
For $\forall n \in \mathbb{N}, w_{n+1} = w_n +b $
The nth-term is: $a + (n)b$
$$w_0 = a$$ $$w_1 = w_0+_1 = w_0 + b = a + b$$ $$w_2 = w_1+_1 = w_1 + b = (a + b) + b = a + 2b$$ $$w_3 = w_2+_1 = w_2 + b = (a + 2b) + b = a + 3b$$ $$....$$ $$w_n = a + (n)b$$
However for $\forall n \in \mathbb{N}, z_n+_1 = z_n + (w_n+_1) x^n$, i don't seem to see the nth-term...
$$z_0 = w_0 = a$$ $$z_1 = z_0+_1 = z_0 + (x_0+_1)x^0 = a + ((a+b)x^0) = a+(a+b)= 2a + b$$ $$z_2 = z_1+_1 = z_1 + (x_1+_1)x^1 = (2a + b) + (a+2b) x^1$$ $$z_3 = z_2+_1 = z_2 + (x_2+_1)x^2 = ((2a + b) + (a+2b) x^1) + (a+3b) x^2 $$ $$....$$ $$z_n = nth-term?$$
Any help would be great appreciated, thanks
Edit: Fixed a typo: $\forall n \in \mathbb{N}, w_n+_1 = w_n +b $
$$z_{n+1} - z_n = w_{n+1}x^n = (a+b(n+1))x^n$$ $$\frac{z_{n+1} - z_n}{z_{n} - z_{n-1}}=x\frac{a+b(n+1)}{a+bn}$$ $$\prod_{k=1}^n\frac{z_{k+1} - z_k}{z_{k} - z_{k-1}}=\frac{z_{n+1} - z_n}{z_{1} - z_{0}}=x^n\prod_{k=1}^n\frac{a+b(k+1)}{a+bk}=x^n\frac{a+b(n+1)}{a+b}$$ $$z_{n+1} - z_n=x^n(z_1-z_0)\frac{a+b(n+1)}{a+b}=x^n(a+b(n+1))$$ $$\sum_{k=0}^{n-1} (z_{k+1} - z_k)=z_n-z_0=\sum_{k=0}^{n-1}x^n(a+b(n+1))$$ $$f(x)=\sum_{k=0}^{n-1}nx^n$$ $$\frac{f(x)}{x}=\sum_{k=0}^{n-1}nx^{n-1}=\left(\sum_{k=0}^{n-1}x^n\right)'=\left(\frac{x^n-1}{x-1}\right)'=\frac{(n-1)x^{n+1}-nx^n+x}{(x-1)^2x}$$ $$\therefore z_n=a+\sum_{k=0}^{n-1}x^n(a+b(n+1))=a+(a+1)\frac{x^n-1}{x-1}+b\frac{(n-1)x^{n+1}-nx^n+x}{(x-1)^2}$$