Problem - 3266
For $n\ge 1$, let $d_n$ denote the length of the line segment connecting the two points where the line $y = x + n + 1$ intersects the parabola $8x^2 = y - \frac{1}{32}$ . Compute the sum $$\sum_{n=1}^{1000}\frac{1}{n\cdot d_n^2}$$
By the quadratic formula, the points of intersection have $x = \frac{1}{16} \pm \frac{1}{4}\sqrt{2n+2}$. The difference between these two
$x$-values is $\frac{1}{2}\sqrt{2n + 2}=\frac{\sqrt{2}}{2}\sqrt{n+1}$.
Because the lines have slope $1$, the length of the line connecting the points is $d_n =\sqrt{n + 1}$. Therefore,
$$\sum_{n=1}^{1000}\frac{1}{n\cdot d_n^2}=\sum_{n=1}^{1000}\frac{1}{n(n+1)}=\sum_{n=1}^{1000}\Big(\frac{1}{n}-\frac{1}{n+1}\Big)=\frac{1}{1}-\frac{1}{1001}=\frac{1000}{1001}$$