Is there a way to write
$\sum_{n=a}^b (Q+P(n-1)) v^n$ in terms of $\sum_{n=a}^b v^n$?
So far I've gotten: (Q-P)$\sum_{n=a}^b v^n$ + P$\sum_{n=a}^b nv^n$.
*I know the last term can be rewritten as $P \frac {(1-v)(av^a-bv^{b+1})+(v^{a+1}-v^{b+1})}{(1-v)^2}$ but not sure where to go from there.
On
The required sum represents the present value of an arithmetically increasing immediate annuity (i.e. a series of $n$ payments made at the end of each period, where the first payment is $Q$ and each additional payment increases by $P$) deferred of $a-1$.
So it is
$$ \sum_{n=a}^b (Q+P(n-1)) v^n=
Q \times\left(_{a-1|}a_{\overline{b^{\phantom{_i}}}\!|}\right)+P\times \left( _{a-1|}(Ia)_{\overline{b^{\phantom{_i}}}\!|}\right)
$$
where
$$_{m|}(Ia)_{\overline{n^{\phantom{_i}}}\!|}=v^{m}(Ia)_{\overline{n^{\phantom{_i}}}\!|},\quad _{m|}a_{\overline{n^{\phantom{_i}}}\!|}=v^m a_{\overline{n^{\phantom{_i}}}\!|}$$
and
$$
(Ia)_{\overline{n^{\phantom{_i}}}\!|}=\frac{\ddot a_{\overline{n^{\phantom{_i}}}\!|}-nv^n}{i},\quad \ddot a_{\overline{n^{\phantom{_i}}}\!|}=(1+i)a_{\overline{n^{\phantom{_i}}}\!|},\quad a_{\overline{n^{\phantom{_i}}}\!|}=\frac{1-v^n}{i},\quad v=\frac{1}{1+i}.
$$
Hint
Starting as you did : $$\sum_{n=a}^b (Q+P(n-1)) v^n=(Q-P)\sum_{n=a}^b v^n+P\sum_{n=a}^b nv^n=(Q-P)\sum_{n=a}^b v^n+Pv\sum_{n=a}^b nv^{n-1}=(Q-P)\sum_{n=a}^b v^n+Pv \frac d {dv}\Big(\sum_{n=a}^b v^{n}\Big)$$ the two summations are simple since $$\sum_{n=a}^b v^{n}=\frac{v^{b+1}-v^a}{v-1}$$