Product of many numbers. Too much!

Based on the definition of $a_n$ , we can calculate the values in the sequence as:

$a_1=\frac{1+3}{1+5}=\frac{4}{6}$

$a_2=\frac{2+3}{2+5}=\frac{5}{7}$

$a_3=\frac{3+3}{3+5}=\frac{6}{8}$

$a_4=\frac{15+3}{15+5}=\frac{18}{20}$ ,

Thus, the product of the first $15$ terms is:

$a_1a_2a_3a_4a_5...a_{12}a_{13}a_{14}a_{15}= \frac{4\times5\times6\times7\times8...15\times16\times17\times18}{6\times7\times8...15\times16\times17\times18\times19\times20}$

Note that both the numerator and denominator have factors of all the consecutive integers between $6$ and $18$ . If both the numerator and denominator have a common factor, then the factor can be eliminated from both numerator and denominator. Thus, the product of the first $15$ terms is:

$a_1a_2a_3a_4a_5...a_{12}a_{13}a_{14}a_{15}= \frac{4\times5\times\color{#D61F06}{6\times7\times8...15\times16\times17\times18}}{\color{#D61F06}{6\times7\times8...15\times16\times17\times18} \color{#333333}{\times19\times20}}=$

$\frac{4\times5}{19\times20}=\boxed{\Large\frac{1}{19}}$

Let

$\begin{aligned} P & = \displaystyle \prod_{n=1}^{15}{\dfrac{n+3}{n+5}} \\ & = \dfrac{{18!}/{3!}}{{20!}/{5!}} \\ & = \dfrac{1}{19\times20}\cdot 4\times5 \\ & = \boxed{\dfrac{1}{19}} \end{aligned}$

$\begin{aligned} P & = a_1a_2a_3...a_{15} \\ & = \frac 46 \cdot \frac 57 \cdot \frac 68 \cdot \frac 79 \cdots \frac {16}{18} \cdot \frac {17}{19} \cdot \frac {18}{20} \\ & = \frac 4{\cancel{6}} \cdot \frac 5{\cancel{7}} \cdot \frac {\cancel{6}}{\cancel{8}} \cdot \frac {\cancel{7}}{\cancel{9}} \cdots \frac {\cancel{16}}{\cancel{18}} \cdot \frac {\cancel{17}}{19} \cdot \frac {\cancel{18}}{20} \\ & = \frac {4 \cdot 5}{19 \cdot 20} \\ & = \boxed{\dfrac 1{19}} \end{aligned}$