Andrews-Schur Identity

$\sum_{k=0}^n q^{k^2+ak}\left[{\matrix{2n-k+a\cr k\cr}}\right]$
$= \sum_{k=-\infty}^\infty q^{10k^2+(4a-1)k}\left[{\matrix{2n+2a+2\cr n-5k\cr}}\right] {[10k+2a+2]\over [2n+2a+2]},\quad$	(1)

where $[x]$ is a Gaussian Polynomial. It is a Polynomial identity for $a=0$, 1 which implies the Rogers-Ramanujan Identities by taking $n\to\infty$ and applying the Jacobi Triple Product identity. A variant of this equation is

$\sum_{k=-\left\lfloor{a/2}\right\rfloor }^n q^{k^2+2ak}\left[{\matrix{n+k+a\cr n-k\cr}}\right]$
$= \sum_{-\left\lfloor{(n+2a+2)/5}\right\rfloor }^{\left\lfloor{n/5}\right\rflo... ...+1)k}\left[{\matrix{2n+2a+2\cr 5-5k\cr}}\right]{[10k+2a+2]\over[2n+2a+2]},\quad$	(2)

where the symbol $\left\lfloor{x}\right\rfloor$ in the Sum limits is the Floor Function (Paule 1994). The Reciprocal of the identity is

$$\sum_{k=0}^\infty {q^{k^2+2ak}\over (q;q)_{2k+a}}=\prod_{j=0}^\infty {1\over(1-q^{2j+1})(1-q^{20j+4a+4})(1-q^{20j-4a+16})}$$ (3)

for $a=0$, 1 (Paule 1994). For $q=1$, (1) and (2) become

$$\sum_{-\left\lfloor{a/2}\right\rfloor }^n{n+k+a\choose n-k}=... ...{n/5}\right\rfloor }{2n+2a+2\choose n-5k} {5k+a+1\over n+a+1}.$$ (4)

References

Andrews, G. E. ``A Polynomial Identity which Implies the Rogers-Ramanujan Identities.'' Scripta Math. 28, 297-305, 1970.

Paule, P. ``Short and Easy Computer Proofs of the Rogers-Ramanujan Identities and of Identities of Similar Type.'' Electronic J. Combinatorics 1, R10 1-9, 1994. http://www.combinatorics.org/Volume_1/volume1.html#R10.