Qザールシュッツの和公式
となる。 z n {\displaystyle z^{n}} の係数を比べると
( a ; q ) n ( b ; q ) n ( q ; q ) n ( c ; q ) n = ∑ m = 0 ∞ ( c a ; q ) m ( c b ; q ) m ( q ; q ) m ( c ; q ) m ( a b c ) m ( a b c ; q ) n − m ( q ; q ) n − m = ∑ m = 0 ∞ ( c a ; q ) m ( c b ; q ) m ( q ; q ) m ( c ; q ) m ( a b c ) m ( a b c ; q ) n ( q 1 + n − m ; q ) m ( q ; q ) n ( a b c q n − m ; q ) m {\displaystyle {\begin{aligned}{\frac {(a;q)_{n}(b;q)_{n}}{(q;q)_{n}(c;q)_{n}}}&=\sum _{m=0}^{\infty }{\frac {({\frac {c}{a}};q)_{m}({\frac {c}{b}};q)_{m}}{(q;q)_{m}(c;q)_{m}}}\left({\frac {ab}{c}}\right)^{m}{\frac {({\frac {ab}{c}};q)_{n-m}}{(q;q)_{n-m}}}\\&=\sum _{m=0}^{\infty }{\frac {({\frac {c}{a}};q)_{m}({\frac {c}{b}};q)_{m}}{(q;q)_{m}(c;q)_{m}}}\left({\frac {ab}{c}}\right)^{m}{\frac {({\frac {ab}{c}};q)_{n}(q^{1+n-m};q)_{m}}{(q;q)_{n}({\frac {ab}{c}}q^{n-m};q)_{m}}}\end{aligned}}}