# Formula:KLS:14.07:16

$\displaystyle {\displaystyle (1-\gamma q^x)(1-\gamma\delta q^x)(1-q^{x-N-1})\dualqHahn{n}@{\mu(x)}{\gamma}{\delta}{N}{q} {}+\gamma q^{x-N-1}(1-q^x)(1-\gamma\delta q^{x+N+1})(1-\delta q^x)\dualqHahn{n}@{\mu(x-1)}{\gamma}{\delta}{N}{q} {}=q^x(1-\gamma)(1-q^{-N-1})(1-\gamma\delta q^{2x})\dualqHahn{n+1}@{\mu(x)}{\gamma q^{-1}}{\delta}{N+1}{q} }$

## Substitution(s)

$\displaystyle {\displaystyle \mu(x)=q^{-x}+\gamma\delta q^{x+1} =q^{-x}+q^{x+\gamma+\delta+1} =q^{-x}+\gamma\delta q^{x+1}}$ &

$\displaystyle {\displaystyle \mu(n)=q^{-n}+\alpha\beta q^{n+1}}$ &
$\displaystyle {\displaystyle \mu(x):=q^{-x}+\gamma\delta q^{x+1}}$ &
$\displaystyle {\displaystyle \mu(x)=q^{-x}+\gamma\delta q^{x+1} =q^{-x}+q^{x+\gamma+\delta+1} =q^{-x}+\gamma\delta q^{x+1}}$ &

$\displaystyle {\displaystyle \mu(n)=q^{-n}+\alpha\beta q^{n+1}}$

## Proof

We ask users to provide proof(s), reference(s) to proof(s), or further clarification on the proof(s) in this space.

## Symbols List

& : logical and
$\displaystyle {\displaystyle R_{n}}$  : dual $\displaystyle {\displaystyle q}$ -Hahn polynomial : http://drmf.wmflabs.org/wiki/Definition:dualqHahn