Answer
$a_{n}= \frac{2n-5}{n(n+1).},\quad n=1,2,3...$
Work Step by Step
Annotate $a_{n}=\displaystyle \frac{A_{n}}{B_{n}}$
The numerators $A_{n}$ form a sequence
$\left[\begin{array}{lllllll}
n & 1 & 2 & 3 & 4 & 5 & ...\\
A_{n} & -3 & -1 & 1 & 3 & 5 & ...
\end{array}\right],$
This is an arithmetic sequence with common difference $d=2$ and initial term $A_{1}=-3$
$A_{1}=-3$
$A_{2}=-3+2$
$A_{3}=-3+2(2)$
$...$
$A_{n}=-3+(n-1)(2)=-3+2n-2=2n-5$
The numerators are done.
Denominator sequence, $B_{n}$:
$\left[\begin{array}{lllllll}
n & 1 & 2 & 3 & 4 & 5 & ...\\
B_{n} & 2 & 6 & 12 & 20 & 30 & ...\\
& 1(2) & 2(3) & 3(4) & 4(5) & 5(6) &
\end{array}\right]\Rightarrow B_{n}=n(n+1)$
Thus, we have:
$a_{n}=\displaystyle \frac{2n-5}{n(n+1).},\quad n=1,2,3...$
