If the sum $\frac{1}{7} + \frac{1\cdot 3}{7\cdot 9} + \frac{1\cdot 3\cdot 5}{7\cdot 9\cdot 11} + ...$ to 20 terms is $\frac{m}{n}$, then $n-4m$?

The following is not particularly clever but works.

Note that $$1\cdot 3 \cdot 5 \cdot \cdots \cdot (2N-1) = \frac{(2N)!}{2^N N!}$$

Using this (or just considering ratios of adjacent terms as suggested in comments), each term in the series is $$\frac{120(2n)! (n+3)!}{n!(2n+6)!} = \frac{15}{(2n+1)(2n+3)(2n+5)} = \frac{15}{8}\left(\frac 1 {2n+1} - \frac 2 {2n+3} + \frac 1 {2n+5} \right)$$

Using telescopic sums, one finds the sum of $N$ terms to be $$\frac 1 4 -\frac{15}8 \left( - \frac 1 {2N+3} + \frac 1 {2N+5}\right)$$ and the answer can be calculated.

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Edit: As Cameron pointed out in the comments, the telescopic nature of the sum is made clearer by realizing you are summing the terms $$\left(\frac 1 3 - \frac 1 5\right) - \left(\frac 1 5 - \frac 1 7\right),\quad \left(\frac 1 5 - \frac 1 7\right) - \left(\frac 1 7 - \frac 1 9\right),\quad \cdots$$

Let's write the last term. What do we notice ? $\dfrac{1\cdot3\cdot5\cdot7\cdot9\cdot11\cdot13\cdot\ldots\cdot39}{7\cdot9\cdot11\cdot\ldots\cdot39\cdot41\cdot43\cdot45}=\dfrac{1\cdot3\cdot5}{41\cdot43\cdot45}$

All terms, starting with the fourth, are of the form $\dfrac{1\cdot3\cdot5}{(2n+1)(2n+3)(2n+5)}$ , which has the

words telescoping series written all over it. :-) Then all that's left to do is add the sum of the first three terms to its result, simplify the fraction, and compute $n-4m$.

Mathematica code for my comment above. I didn't make it an answer because it is only a programming hack, and doesn't explain the mathematical underpinnings.

c = Total[
    Join[{(a = 1)/(b = 7)}, 
         Table[(a *= j)/(b *= (j + 6)), {j, 3, 39, 2}]]];
Denominator[c] - 4 Numerator[c]