An identity on $\small{}_pF_q\left(\left.\begin{array}{c} a_1+1,a_2+1,\dots ,a_p+1\\ b_1+1,b_2+1,\dots ,b_q+1\end{array}\right| z\right)$ I stumbled upon this relation while trying to answer this post. I was trying to find a relation between the two generalized hypergeometric functions,
$$A=\,_3F_2\left(\color{blue}{\tfrac12,\tfrac12},\tfrac12;\color{red}{\tfrac32,\tfrac32};\color{fuchsia}{\tfrac12}\right)$$
$$B=\,_3F_2\left(\tfrac32,\tfrac32,\tfrac32;\tfrac52,\tfrac52;\tfrac12\right)$$
It seems,
$$A+\tfrac1{18}B = \,_2F_1\left(\tfrac12,\tfrac12;\tfrac32;\tfrac12\right) =\frac{\pi}{2\sqrt2}$$
Note that from a $_3F_2$, the sum reduces to a $_2F_1$, and  $\tfrac1{18}= \color{blue}{\tfrac12\tfrac12} \color{red}{\tfrac23\tfrac23} \color{fuchsia}{\tfrac12} $.

Question: In general, let
$$p=q+1\\c_n = a_n+1\\d_n = b_n+1$$ 
where $a_n, b_n$ are arbitrary but the last pair must satisty $a_p+1=b_q$. Is it true that,
$$
{}_pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q  \end{array}\right| z\right)+z\,\frac{a_1a_2\dots a_{p-1}}{b_1b_2\dots b_q}{}_pF_q\left(\left.\begin{array}{c} c_1,c_2,\dots ,c_p\\ d_1,d_2,\dots ,d_q  \end{array}\right| z\right)\\={}_{p-1}F_{q-1}\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_{p-1}\\ b_1,b_2,\dots ,b_{q-1}  \end{array}\right| z\right)\\
{}
\\
$$
(Note: The pair $a_p,b_q$ disappears in the $\text{RHS}$.)
 A: We first use the differentiation formula for the generalized hypergeometric function
\begin{equation}
 \frac{a_1a_2\dots a_{p}}{b_1b_2\dots b_q}{}_pF_q\left(\left.\begin{array}{c} c_1,c_2,\dots ,c_p\\ d_1,d_2,\dots ,d_q  \end{array}\right| z\right)=\frac{d}{dz}{}_pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q  \end{array}\right| z\right)
\end{equation} 
Then, the LHS of the proposed identity can be written as
\begin{equation}
 _pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q  \end{array}\right| z\right)+z\,\frac{a_1a_2\dots a_{p-1}}{b_1b_2\dots b_q}{}_pF_q\left(\left.\begin{array}{c} c_1,c_2,\dots ,c_p\\ d_1,d_2,\dots ,d_q  \end{array}\right| z\right)=\left( 1+\frac{z}{a_p}\frac{d}{dz} \right){} _pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q  \end{array}\right| z\right)\tag{1}\label{eq1}
\end{equation}
To differentiate the hypergeometric function, we use the Euler's integral transform
\begin{align}
& _pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q  \end{array}\right| z\right)\\
&=\frac{\Gamma(b_q)}{\Gamma(a_p)\Gamma(b_q-b_p)} \int_0^1t^{a_p-1}\left( 1-t \right)^{b_q-a_p-1}{}_{p-1}F_{q-1}\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_{p-1}\\ b_1,b_2,\dots ,b_{q-1}  \end{array}\right| t\right)\,dt
\end{align}
Here $b_q=a_p+1$, then
\begin{align}
  _pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q  \end{array}\right| z\right)&=
  a_p \int_0^1t^{a_p-1}{}_{p-1}F_{q-1}\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_{p-1}\\ b_1,b_2,\dots ,b_{q-1}  \end{array}\right| zt\right)\,dt\\
  &=\frac{a_p}{z^{a_p}} \int_0^zu^{a_p-1}{}_{p-1}F_{q-1}\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_{p-1}\\ b_1,b_2,\dots ,b_{q-1}  \end{array}\right| u\right)\,du
\end{align}
Then
\begin{align}
 \frac{d}{dz}&\,{} _pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q  \end{array}\right| z\right)\\
&=\frac{a_p}{z}\,{}_{p-1}F_{q-1}\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_{p-1}\\ b_1,b_2,\dots ,b_{q-1}  \end{array}\right| z\right)-\frac{a_p}{z} \,{}_pF_q\left(\left.\begin{array}{c} a_1,a_2,\dots ,a_p\\ b_1,b_2,\dots ,b_q  \end{array}\right| z\right)
\end{align} 
Plugging this expression in eq. \eqref{eq1} we find theRHS of the proposed identity.
