Weird form

The given expression is $\frac{n(n+1)}{(n-1)^3}$ . If we multiply by $n^3$ , then we are looking for the digit before the base point. For $n=11$ we have $\lfloor\frac{n^4(n+1)}{(n-1)^3}\rfloor=175=11^2+4*11+10$ , so that the answer is $\boxed{10}$ .

Notice that the given expression is a well-known expression from calculus and the theory of generating functions; namely, it generates the square numbers like so:

$\frac{\frac{1}{n}(\frac{1}{n}+1)}{(1-\frac{1}{n})^3} = \sum_{k=1}^{\infty} \frac{k^2}{n^k}$

Since this translates to a (more or less) direct base $n$ expansion of the expression, all we need to do is look at the third digit, which happens to be $k^2 = 3^2 = 9$ .

However, notice that at $k = 4$ , $k^2 = 4^2 = 16$ exceeds $n = 11$ . Thus, we need to carry over, resulting in a 10 instead.

Therefore, the third digit is $\boxed{10}$ .