Question

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

Answer 1

`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`