Abel's binomial theorem Abel's generalization of the binomial theorem goes as follows
$$(a+x)^n=x^n+ \sum_{i=1}^n \binom{n}{i} a(a+iz)^{i-1}(x-iz)^{n-i}$$
The objective is to prove the generalization by induction. My first step following the textbook was to set $n=k$ and integrate both sides with respect to $x$, in order to obtain; $$\frac {(a+x)^{k+1}}{k+1}=\frac {x^{k+1}}{k+1}+ \sum_{i=1}^k \binom{k}{i} \frac {a(a+iz)^{i-1}(x-iz)^{k-i+1}}{k-i+1} + C$$
The next step was to find the value of the integration constant $C$ by setting $x=(k+1)z$ in the formula above and use the inductive hypothesis. This is the part where I got stuck and cannot seem to figure it out...
$$\frac {(a+z(k+1))^{k+1}}{k+1}=\frac {(z(k+1))^{k+1}}{k+1}+ \sum_{i=1}^k \binom{k}{i} \frac {a(a+iz)^{i-1}(z(k+1)-iz)^{k-i+1}}{k-i+1} + C$$
I see how multiplying both sides by $k+1$ would give us $$(a+z(k+1))^{k+1}= (z(k+1))^{k+1}+\sum_{i=1}^k \binom{k+1}{i} a(a+iz)^{i-1}(z(k+1)-iz)^{k-i+1}+ C$$
where, again, $(k+1)z=x$. But I still am confused as to determining $C$. It was previously tried by setting $z=0$, that is $C(a,0)$ for $C(a,z)$, where $a,z \in \mathbb {R}$. But this computes only for $z$, and therefore the answer could not hold...
 A: Here we complete OPs proof by induction. Note, we will employ the induction hypothesis twice. Assuming $a\ne 0$ we can write OPs version of Abel's binomial theorem somewhat more convenient as:
\begin{align*}
(a+x)^k&=\sum_{q=0}^k\binom{k}{q}a(a+qz)^{q-1}(x-qz)^{k-q}\tag{1.1}
\end{align*}
The induction step we want to show is
Induction step:
\begin{align*}
(a+x)^{k+1}&=\sum_{q=0}^{k+1}\binom{k+1}{q}a(a+qz)^{q-1}(x-qz)^{k+1-q}\tag{1.2}
\end{align*}
OP already integrated (1.1) and obtained after some simplification
\begin{align*}
(a+x)^{k+1}&=\sum_{q=0}^k\binom{k+1}{q}a(a+qz)^{q-1}(x-qz)^{k+1-q}\tag{2}\\
&\qquad+(k+1)C(a,z)
\end{align*}
with $C(a,z)$ an integration constant dependent on the constants $a$ and $z$    . Comparison of (2) with (1.2)  shows that $(k+1)C(a,z)$ is the summand with index $q=k+1$, so that we have to show
\begin{align*}
\color{blue}{C(a,z)=\frac{a\left(a+(k+1)z\right)^k}{k+1}}\tag{3}
\end{align*}
The representation of $C(a,z)$ makes it plausible, that we start using the induction hypothesis (1.1) by letting $x=(k+1)z$. We multiply (1.1) also with $(a+(k+1)z)$ and obtain
\begin{align*}
&(a+(k+1)z)^{k+1}\\
&\qquad=(a+(k+1)z)\sum_{q=0}^k\binom{k}{q}a(a+qz)^{q-1}((k+1-q)z)^{k-q}\\
&\qquad=\sum_{q=0}^k\binom{k}{q}a(a+qz)^{q}((k+1-q)z)^{k-q}\\
&\qquad\qquad+\sum_{q=0}^k\binom{k}{q}a(a+qz)^{q-1}((k+1-q)z)^{k+1-q}\tag{4.1}\\
\end{align*}
where we use $a+(k+1)z=(a+qz)+((k+1-q)z)$. On the other hand we obtain from (2) by using $\binom{k+1}{q}=\binom{k}{q}+\binom{k}{q-1}$
\begin{align*}
&(a+(k+1)z)^{k+1}\\
&\qquad=\sum_{q=0}^k\binom{k+1}{q}a(a+qz)^{q-1}(x-qz)^{k+1-q}\\
&\qquad=\sum_{q=0}^k\binom{k}{q}a(a+qz)^{q-1}((k+1-q)z)^{k+1-q}\\
&\qquad\qquad+\sum_{q=0}^k\binom{k}{q-1}a(a+qz)^{q-1}((k+1-q)z)^{k+1-q}\\
&\qquad\qquad+(k+1)C(a,z)\tag{4.2}
\end{align*}

Equating (4.1) and (4.2) one of the sums cancels and we get after rearranging
\begin{align*}
\color{blue}{(k+1)}&\color{blue}{C(a,z)}=\sum_{q=0}^k\binom{k}{q}a(a+qz)^q((k+1-q)z)^{k-q}\\
&\qquad-\sum_{q=0}^{k}\binom{k}{q-1}a(a+qz)^{q-1}((k+1-q)z)^{k+1-q}\\
&=\sum_{q=0}^k\left(\binom{k}{q}-\binom{k}{q-1}\frac{(k+1-q)z}{a+qz}\right)
a(a+qz)^q((k+1-q)z)^{k-q}\\
&=\sum_{q=0}^k\left(\binom{k}{q}-\binom{k}{q}\frac{qz}{a+qz}\right)
a(a+qz)^q((k+1-q)z)^{k-q}\\
&=\sum_{q=0}^k\binom{k}{q}\left(1-\frac{qz}{a+qz}\right)
a(a+qz)^q((k+1-q)z)^{k-q}\\
&=a\sum_{q=0}^k\binom{k}{q}a(a+qz)^{q-1}((k+1-q)z)^{k-q}\\
&\,\,\color{blue}{=a(a+(k+1)z)^k}
\end{align*}
and the claim (3) and so the induction step (1.2) follows. Note, in the last line we again used the induction hypothesis (1.1) with $x=(k+1)z$.

