- #1
physicsjock
- 89
- 0
Hey,
I've been trying to find a way to simplify these two series coefficients,
[itex]\begin{align}
& {{a}_{4n}}={{a}_{o}}\frac{2!6!...(n-6)!(n-2)}{4!8!...(n-4)!n!} \\
& {{a}_{4n+1}}={{a}_{1}}\frac{3!7!...(n-6)!(n-2)}{5!9!...(n-4)!n!} \\
& E.g. \\
& {{a}_{12}}={{a}_{o}}\frac{2!6!10!}{4!8!12!}\,= \frac{{{a}_{o}}}{3\cdot 4\cdot 7\cdot 8\cdot 11\cdot 12}\,\,\,\,\,\,and \\
& {{a}_{13}}={{a}_{1}}\frac{3!7!11!}{5!9!13!}\,=\,\,\frac{{{a}_{1}}}{4\cdot 5\cdot 8\cdot 9\cdot 12\cdot 13} \\
\end{align}
[/itex]
This is the best way I could find but I was wondering if there were any ways to avoid the use of ...'s
Is there any way to simplify these coefficients?
Thanks in advance
I've been trying to find a way to simplify these two series coefficients,
[itex]\begin{align}
& {{a}_{4n}}={{a}_{o}}\frac{2!6!...(n-6)!(n-2)}{4!8!...(n-4)!n!} \\
& {{a}_{4n+1}}={{a}_{1}}\frac{3!7!...(n-6)!(n-2)}{5!9!...(n-4)!n!} \\
& E.g. \\
& {{a}_{12}}={{a}_{o}}\frac{2!6!10!}{4!8!12!}\,= \frac{{{a}_{o}}}{3\cdot 4\cdot 7\cdot 8\cdot 11\cdot 12}\,\,\,\,\,\,and \\
& {{a}_{13}}={{a}_{1}}\frac{3!7!11!}{5!9!13!}\,=\,\,\frac{{{a}_{1}}}{4\cdot 5\cdot 8\cdot 9\cdot 12\cdot 13} \\
\end{align}
[/itex]
This is the best way I could find but I was wondering if there were any ways to avoid the use of ...'s
Is there any way to simplify these coefficients?
Thanks in advance