Generating a rate equation from a paper I'm going through equations in this paper Structure of Growing Networks with Preferential Linking, I was not able to understand how they derived equation $[3]$ by summing up equation $[2]$. 
eqn [2]
$$p(q,s,t+1)=\sum_{l=0}^{m}\binom{m}{l}\left [\frac{ q-l+am}{(1+a)mt} \right ]^{l}\left [1-\frac{ q-l+am}{(1+a)mt} \right ]^{m-l}p(q-l,s,t)$$ 
substituting for sum over $p(q,u,t)$ using:
$P(q,t)=\sum_{u=1}^{t}p(q,u,t)/t$
They obtain $eqn [3]$
$$(t+1)P(q,t+1)-p(q,t+1,t+1)=(1-\frac{q+am}{1+a})P(q,t)+(\frac{q-1+am}{1+a})P(q-1,t)+\mathcal{O}(\frac{P}{t})$$
My understanding is that $P(q,t+1)$ is the probability of having sites with $q$ in_links, this consists of two parts, one being the set of sites with in_degree $q-1$ at time $t$ that gain one additional link in time step $t+1$. Another set is due to loosing sites that already have $q$ in_links at time $t$ and gain a new links, thus no longer contribute to $P(q,t+1)$, therefore have a negative sign. These should correspond the second and first term on the right hand side of the equation $[3]$. What I don't understand is how to obtain the multiplying factor for each term. 
I would be very grateful for any insight. 
 A: For the LHS of eqn. 2, observing that
$$
\sum_{s=1}^{t}p(q,s,t+1)=\sum_{s=1}^{t+1}p(q,s,t+1)-p(q,t+1,t+1)
$$
so that
$$
\sum_{s=1}^{t}p(q,s,t+1)=(t+1)\sum_{s=1}^{t+1}\frac{p(q,s,t+1)}{t+1}-p(q,t+1,t+1)
$$
and using $P(q,t)=\sum_{s=1}^t \frac{p(q,s,t)}{t}$, we have
$$
LHS=(t+1)P(q,t+1)-p(q,t+1,t+1)
$$
For the RHS, observe that we neglect the terms of order higher than $1/t$ so that you have to sum the first two terms for $l=0, 1$. So, calling $b(l)=\frac{ q-l+am}{(1+a)mt}$, we have
$$\sum_{s=1}^t \sum_{l=0}^{m}\binom{m}{l}b^{l}(1-b)^{m-l}p(q-l,s,t)=
\sum_{l=0}^{m}\binom{m}{l}b^{l}(1-b)^{m-l}\underbrace{\sum_{s=1}^t t\frac{p(q-l,s,t)}{t}}_{tP(q-l,t)}$$ 
and considering the first two term only, we obtain:
$$
RHS=(1-b(0))^m tP(q,t)+m b(1)(1-b(1))^{m-1}tP(q-1,t)+\mathcal{O}(P/t)
$$
Using the binomial approximation $(1+x)^n\approx 1+nx$ we obtain:
$$
(1-b(0))^m \approx 1-m\frac{q-l+am}{(1+a)mt}=1-\frac{q-l+am}{(1+a)t}
$$
and 
$$
m b(1)(1-b(0))^m \approx m b(1)=m\frac{q-l+am}{(1+a)mt}=\frac{q-l+am}{(1+a)t}
$$
Putting all togheter we have
$$
RHS=\left(1-\frac{q-l+am}{(1+a)t}\right) t P(q,t)+\frac{q-l+am}{(1+a)t}tP(q-1,t)+\mathcal{O}(P/t)
$$
and finally
$$
RHS=\left(t-\frac{q-l+am}{(1+a)}\right) P(q,t)+\frac{q-l+am}{(1+a)}P(q-1,t)+\mathcal{O}(P/t).
$$
The LHS and RHS give us the eqn. 3
$$
(t+1)P(q,t+1)-p(q,t+1,t+1)=\left(t-\tfrac{q-l+am}{(1+a)}\right) P(q,t)+\tfrac{q-l+am}{(1+a)}P(q-1,t)+\mathcal{O}(P/t).
$$
