1983
Problem - 3318
Let $\alpha_n$ and $\beta_n$ be two roots of equation $x^2+(2n+1)x+n^2=0$ where $n$ is a positive integer. Evaluate the following expression $$\frac{1}{(\alpha_3+1)(\beta_3+1)}+\frac{1}{(\alpha_4+1)(\beta_4+1)}+\cdots+\frac{1}{(\alpha_{20}+1)(\beta_{20}+1)}$$
By Vieta's theorem:
$$\alpha_n + \beta_n = -(2n+1)\quad\text{and}\quad \alpha_n\cdot\beta_n=n^2$$
Therefore $$(\alpha_n+1)(\beta_n+1)=\alpha_n\cdot\beta_n + (\alpha_n+\beta_n)+1 = n^2 -2(n+1)+1=(n-2)n$$
It follows
\begin{align*}
&\frac{1}{(\alpha_3+1)(\beta_3+1)}+\frac{1}{(\alpha_4+1)(\beta_4+1)}+\cdots+\frac{1}{(\alpha_{20}+1)(\beta_{20}+1)}\\
=&\frac{1}{1\times 3} + \frac{1}{2\times 4} +\frac{1}{3\times 5} +\cdots + \frac{1}{18\times 20}\\
=&\frac{1}{2}\times\Big(\frac{1}{1}-\frac{1}{3}\Big)+\frac{1}{2}\times\Big(\frac{1}{2}-\frac{1}{4}\Big)+\frac{1}{2}\times\Big(\frac{1}{3}-\frac{1}{5}\Big)+\cdots +\frac{1}{2}\times\Big(\frac{1}{18}-\frac{1}{20}\Big)\\
=&\frac{1}{2}\times\Big(\frac{1}{1}+\frac{1}{2}-\frac{1}{19}-\frac{1}{20}\Big)\\
=&\boxed{\frac{531}{760}}
\end{align*}