2. Find the sum $a_{1}+a_{2}+\ldots+a_{1985}$, if

$$
a_{k-1}=\frac{3 k^{2}-3 k+1}{\left(k^{2}-k\right)^{3}} \text { for } k \geq 2
$$
Solution. For $k \geq 2$ we have

$$
a_{k-1}=\frac{3 k^{2}-3 k+1}{\left(k^{2}-k\right)^{3}}=\frac{k^{3}-(k-1)^{3}}{k^{3}(k-1)^{3}}=\frac{1}{(k-1)^{3}}-\frac{1}{k^{3}}=b_{k-1}-b_{k}
$$

where $b_{n}=\frac{1}{n^{3}}$ for $n \geq 1$. Then

$$
\begin{aligned}
a_{1}+a_{2}+\ldots+a_{1985} & =b_{1}-b_{2}+b_{2}-b_{3}+\ldots+b_{1984}-b_{1985}+b_{1985}-b_{1986} \\
& =b_{1}-b_{1985} \\
& =1-\frac{1}{1986^{3}}=\frac{1986^{3}-1}{1986^{3}}
\end{aligned}
$$