下面给出了一个数列的前五项。 $\frac{3}{2}\quad \frac{7}{8}\quad \frac{11}{18}\quad \frac{15}{32}\quad \frac{19}{50}$

1. Identify and extend the numerator pattern.

The numerators are
$3,\ 7,\ 11,\ 15,\ 19$
They increase by $4$ each time, so the next two numerators are
$19+4=23,\qquad 23+4=27.$

2. Identify and extend the denominator pattern and write the next two terms.

The denominators are
$2,\ 8,\ 18,\ 32,\ 50$
These are
$2(1^2),\ 2(2^2),\ 2(3^2),\ 2(4^2),\ 2(5^2).$
So the next two denominators are
$2(6^2)=72,\qquad 2(7^2)=98.$
Hence the next two terms are
$\frac{23}{72},\qquad \frac{27}{98}.$

1. Match the given numerator and denominator to the sequence pattern.

The numerators are $3,7,11,15,19,\ldots$, increasing by $4$ each time.

So the numerator of the $n$th term is $4n-1$.

For the given term $\dfrac{47}{288}$,
$4k-1=47$
$4k=48$
$k=12$

Check denominator pattern: $2,8,18,32,50,\ldots=2n^2$.

When $k=12$, denominator $=2(12)^2=288$, so this matches.

1. Find the numerator pattern in terms of $n$.

The numerators are
$3,\ 7,\ 11,\ 15,\ 19,\ldots$

This is an arithmetic sequence increasing by $4$ each time.

For the $n$th term, the numerator is
$3+4(n-1)=4n-1.$

2. Find the denominator pattern in terms of $n$ and form the nth term.

The denominators are
$2,\ 8,\ 18,\ 32,\ 50,\ldots$

These can be written as
$2(1^2),\ 2(2^2),\ 2(3^2),\ 2(4^2),\ 2(5^2),\ldots$

So the denominator of the $n$th term is $2n^2$.

Hence the $n$th term is
$\frac{4n-1}{2n^2}.$