# Formula:KLS:14.07:04

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

## Substitution(s)

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

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

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

## 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