Spider web in the unit square

Someone who can create and embed pictures should post a solution, but here's one for now.

The total area is the area of $n$ triangles, each of whose bases is one of the $n$ subintervals on the $x$ -axis, whose top vertex is the intersection of the segments $\overline{A_\ell B_\ell}$ and $\overline{A_{\ell+1}B_{\ell+1}}.$ Some simple algebra shows that the $y$ -coordinate of that intersection point is $\frac{k(k+1)}{n(n+1)},$ where $k = n+1-\ell,$ so the sum is $\begin{aligned} S_n &= \frac1{2n} \sum_{k=1}^n \frac{k(k+1)}{n(n+1)} \\ &= \frac1{2n^2(n+1)} \sum_{k=1}^n k(k+1) \\ &= \frac1{2n^2(n+1)} \left( \frac{n(n+1)(2n+1)}6 + \frac{n(n+1)}2 \right) \\ &= \frac{2n+1}{12n} + \frac1{4n} = \frac{n+2}{6n}. \end{aligned}$ So the answer is $1+2+6+0 = \fbox{9}.$

First check the situation for $n=2$ . For $n=3, S_n=\frac{1}{3}$ . So, $2*a+b=1k; 2*c+d=3k$ as $n=2$ ......... (1)

Now if $n\rightarrow \infty$ the united area is looks to be cut by a hyperbola and have the area $S_{\infty}=\frac{1}{6}$ .

See, $S_{\infty}=\lim \limits_{n \to\infty} \frac{an+b}{cn+d}=\frac{a}{c}$ .

Hence, $a=p,c=6p$ . And from ( 1 ) we get $\frac{2*p+b}{2*6p+d}=\frac{1k}{3k}=\frac{4k}{12k}$ . Comparing this we get $b=2p$ . $k=p$ .

Now we check the case for $n=3$ . Here $S_3=\frac{5}{18}=\frac{3*p+2p}{3*6p+d}$ . After seeing both case we can conclude that $d=0$ . Hence, $S_n=\frac{an+b}{cn}$ .

So, in all these cases $p$ has actually no effect. So, take the most convenient case when p=1. Hence, $a=1,b=2,c=6,d=0$ . So, $a+b+c+d=9$ .