How do you find #a_8# given #a_n=(n!)/(2n)#?

1 Answer
Alan P.
Nov 1, 2017

#a_8=color(red)(2520)#

Explanation:

If #a_color(blue)n=(color(blue)n!)/(2color(blue)n)#

then #a_color(blue)8=(color(blue)8!)/(2 * color(blue)8)#

#color(white)("XXX")=(8xx7xx6xx5xx4xx3xx2xx1)/(2xx8)#

#color(white)("XXX")=7xx6xx5xx4xx3#

#color(white)("XXX")=42xx5xx4xx3#

#color(white)("XXX")=210xx4xx3#

#color(white)("XXX")=840xx3#

#color(white)("XXX")=2520#

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