The recurrence equation is written as $ U_k = \sum_{m=0}^{k-1} a_{k-1-m} U_m \tag{1} $ Let's form the generating function $f(x) = \sum_{k=0}^\infty x^k U_k$. Then, summing eq. 1, multiplied on both sides with $x^k$, from $k=1$ to $\infty$: $ f(x) - 1 = x \cdot f(x) \cdot \sum_{k=0}^\infty a_k x^k = x f(x) g(x) $ Since $a_k = a_0 + (a_1-a_0) k$, $g(x) = \sum_{k=0}^\infty a_k x^k = a_0 \frac{1-2x}{(1-x)^2} + a_1 \frac{x}{(1-x)^2}$. Hence
$ f(x) = \frac{1}{1-x g(x)} = \frac{(1-x)^2}{(1-x)^2 - a_0 x + x^2 (2a_0 - a_1)} $
The case of $a_n = n+1$, that corresponds to $a_0=1$ and $a_1 = 2$, we recover the generating function of $F_{2k}$: $ f(x) = \frac{(1-x)^2}{1-3 x + x^2} = 1 + x + 3 x^2 + 8 x^3 + \mathcal{o}(x^3) $
The case of $a_n = b n -1$ maps to $a_0 =-1$ and $a_1 = b-1$ and generating function: $ f(x) = \frac{(1-x)^2}{1-x- b x^2} $ The above generating function does not correspond to $U_k = b^2 (b+1)^{k-2}$ for $k \geqslant 2$.
In[44]:= u[0] = 1; In[45]:= u[k_Integer?Positive] := u[k] = Expand[Sum[a[k - 1 - m] u[m], {m, 0, k - 1}]] In[49]:= (Sum[u[k] x^k, {k, 0, 12}] /. {a[m_] :> b m - 1}) - Series[-((-1 + x)^2/(-1 + x + b x^2)), {x, 0, 12}] Out[49]= SeriesData[x, 0, {}, 13, 13, 1]
The asymptotic growth of $U_k$ is determined by poles of $f(x)$.