<s>[INST] write formula to predict nth term of sequence $1, 1\cdot3, 1\cdot3\cdot5, 1\cdot3\cdot5\cdot(2n-1)$ How can I write a formula for a sequence with the following behavior:
{$1, 1\cdot3, 1\cdot3\cdot5, 1\cdot3\cdot5\cdot7, 1\cdot3\cdot5\cdot7\cdot9$}
1st term is $1$
2nd term is $1 \cdot 3 = 3$
3rd term is $1 \cdot 3 \cdot 5 = 15$
4th term is $1 \cdot 3 \cdot 5 \cdot 7 = 105$
Please show steps on how to arrive at the answer:
Ted
 [/INST] The classic trick is to multiply and divide by the even terms:
$$
a_n=1\cdot3\cdot\dots\cdot(2n-1)=\frac{1\cdot2\cdot3\cdot \dots\cdot 2n}{2\cdot 4\cdot (2n)}
$$
The numerator is $(2n)!$. As for the denominator, we can factor each of the $n$ terms by $2$, so the denominator is $2^nn!$.
Finally, $a_n=\frac{(2n)!}{2^nn!}$ 
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