Answer
$a_{n}= \frac{2n-1}{n^{2}}$
Or:
$a_{n}= \frac{1+2(n-1)}{n^{2}}$
Work Step by Step
We are given:
$\frac{1}{1}, \frac{3}{4}, \frac{5}{9}, \frac{7}{16}, \frac{9}{25}$ ...
We notice that the numerators increase by $2$ between terms. We also see that the denominators are powers of integers ($1^2$, $2^2$, $3^2$, etc). We find the pattern:
$a_{1}= \frac{2(1)-1}{1^{2}}$
$a_{2}= \frac{2(2)-1}{2^{2}}$
$a_{3}= \frac{2(3)-1}{3^{2}}$
$a_{4}= \frac{2(4)-1}{4^{2}}$
$a_{5}= \frac{2(5)-1}{5^{2}}$
Therefore:
$a_{n}= \frac{2n-1}{n^{2}}$
Or:
$a_{n}= \frac{1+2(n-1)}{n^{2}}$
