Can this be a series???

Checking the quotient of each two successive terms:

$6 \div 1 = 6$

$60 \div 6 = 10$

$840 \div 60 = 14$

We see that these are in an arithmetic progression (formally, $a_{n+1}\div a_n=2+4n$ ); following this we get that the quotient of the 10th and 9th term is $\boxed{38}$ .

$\begin{array} {llll} a_1 & = 1 & & \\ a_2 & = 6 & = 6a_1 & = [6+4(2-2)]a_1 \\ a_3 & = 60 & = 10a_2 & = [6+4(3-2)]a_2 \\ a_4 & = 840 & = 14a_3 & = [6+4(4-2)]a_3 \\ ... \\ a_n & & & = [6+4(n-2)]a_{n-1} \end{array}$

Therefore,

$\dfrac {a_{10}}{a_9} = \dfrac {[6+4(10-2)]a_9}{a_9} = 6+4(8) = \boxed{38}$

F(n) = (2n-1)!/(n-1)!

So F(10) = 19!/9!

And F(9) = 17!/8!

F(10)/F(9) = (19!/9!)(8!/17!) = 38

(n + 1) th term/ n th term = n th term in the arithmetic sequence (6, 10, 14, 18, ...... )

Then

10 th term/9 th term = 9 th term in the arithmetic sequence (6, 10, 14, 18, ...... ) =

6 + (9 - 1)(4) = 38

Hint : Check their divisors .