Formula:KLS:14.02:05

{\displaystyle{\displaystyle{\displaystyle\sum_{x=0}^{N}\frac{\qPochhammer{% \alpha q,\beta\delta q,\gamma q,\gamma\delta q}{q}{x}}{\qPochhammer{q,\alpha^{% -1}\gamma\delta q,\beta^{-1}\gamma q,\delta q}{q}{x}}{}\frac{(1-\gamma\delta q% ^{2x+1})}{(\alpha\beta q)^{x}(1-\gamma\delta q)}\qRacah{m}@@{\mu(x)}{\alpha}{% \beta}{\gamma}{\delta}{q}\qRacah{n}@@{\mu(x)}{\alpha}{\beta}{\gamma}{\delta}{q% }=h_{n}\,\Kronecker{m}{n}}}}

Substitution(s)

{\displaystyle{\displaystyle{\displaystyle\mu(x)=q^{-x}+\gamma\delta q^{x+1}=% \lambda(x)=q^{-x}+cq^{x-N}=q^{-x}+q^{x+\gamma+\delta+1}=2a\cos\theta}}}

&

${\displaystyle{\displaystyle{\displaystyle\lambda(x)=x(x+\gamma+\delta+1)}}}$ &
Failed to parse (LaTeXML (experimental; uses MathML): Invalid response ("") from server "http://latexml:8080/convert/":): {\displaystyle {\displaystyle h_n=\frac{\qPochhammer{\alpha^{-1}\beta^{-1}\gamma,\alpha^{-1}\delta,\beta^{-1},\gamma\delta q^2}{q}{\infty}} {\qPochhammer{\alpha^{-1}\beta^{-1}q^{-1},\alpha^{-1}\gamma\delta q,\beta^{-1}\gamma q,\delta q}{q}{\infty}} {}\frac{(1-\alpha\beta q)(\gamma\delta q)^n}{(1-\alpha\beta q^{2n+1})} \frac{\qPochhammer{q,\alpha\beta\gamma^{-1}q,\alpha\delta^{-1}q,\beta q}{q}{n}}{\qPochhammer{\alpha q,\alpha\beta q,\beta\delta q,\gamma q}{q}{n}} =\left\{\begin{array}{ll} \displaystyle\frac{\qPochhammer{\beta^{-1},\gamma\delta q^2}{q}{N}}{\qPochhammer{\beta^{-1}\gamma q,\delta q}{q}{N}} \frac{(1-\beta q^{-N})(\gamma\delta q)^n}{(1-\beta q^{2n-N})} \frac{\qPochhammer{q,\beta q,\beta\gamma^{-1}q^{-N},\delta^{-1}q^{-N}}{q}{n}}{\qPochhammer{\beta q^{-N},\beta\delta q,\gamma q,q^{-N}}{q}{n}} &<br /> \quad\textrm{if}\quad\alpha q=q^{-N}\\ \\ \displaystyle\frac{\qPochhammer{\alpha\beta q^2,\beta\gamma^{-1}}{q}{N}}{\qPochhammer{\alpha\beta\gamma^{-1}q,\beta q}{q}{N}} \frac{(1-\alpha\beta q)(\beta^{-1}\gamma q^{-N})^n}{(1-\alpha\beta q^{2n+1})} \frac{\qPochhammer{q,\alpha\beta q^{N+2},\alpha\beta\gamma^{-1}q,\beta q}{q}{n}}{\qPochhammer{\alpha q,\alpha\beta q,\gamma q,q^{-N}}{q}{n}} &<br /> \quad\textrm{if}\quad\beta\delta q=q^{-N}\\ \\ \displaystyle\frac{\qPochhammer{\alpha\beta q^2,\delta^{-1}}{q}{N}}{\qPochhammer{\alpha\delta^{-1}q,\beta q}{q}{N}} \frac{(1-\alpha\beta q)(\delta q^{-N})^n}{(1-\alpha\beta q^{2n+1})} \frac{\qPochhammer{q,\alpha\beta q^{N+2},\alpha\delta^{-1}q,\beta q}{q}{n}} {\qPochhammer{\alpha q,\alpha\beta q,\beta\delta q,q^{-N}}{q}{n}} &<br /> \quad\textrm{if}\quad\gamma q=q^{-N} \end{array}\right.}} } &
${\displaystyle{\displaystyle{\displaystyle\mu(x):=q^{-x}+\gamma\delta q^{x+1}}}}$ &
${\displaystyle{\displaystyle{\displaystyle\mu(x)=q^{-x}+\gamma\delta q^{x+1}=% \lambda(x)=q^{-x}+cq^{x-N}=q^{-x}+q^{x+\gamma+\delta+1}=2a\cos\theta}}}$ &