A really large staircase

Consider a side view of the staircase. We see that the area of this cross-section is the sum of the areas of infinitely many $45-45-90$ triangles, with leg lengths $1,$ $\dfrac12,$ $\dfrac13,$ $\dfrac14, \ldots.$

In general, the area of a 45-45-90 triangle with leg length $\dfrac1k$ is $\dfrac12 \cdot \left(\dfrac1k\right)^2 = \dfrac12 \cdot \dfrac{1}{k^2}.$ So, the sum of the areas that we want is given by

$\displaystyle\sum_{k=1}^\infty \dfrac12 \cdot \dfrac{1}{k^2}.$

The crucial step here is to recall the famous infinite sum $\displaystyle\sum_{k=1}^\infty \dfrac{1}{k^2} = \dfrac{\pi^2}{6},$ giving us

$\displaystyle\sum_{k=1}^\infty \dfrac12 \cdot \dfrac{1}{k^2} = \dfrac12 \displaystyle\sum_{k=1}^\infty \dfrac{1}{k^2} = \dfrac12 \cdot \dfrac{\pi^2}{6} = \dfrac{\pi^2}{12}.$

Since the width of the staircase is $3,$ the total volume is thus $3 \cdot \dfrac{\pi^2}{12} = \dfrac{\pi^2}{4} \approx 2.467,$ so the answer is $\boxed{246}.$

In the following, we work in feet (the unit, not the part of the body which we use to walk up stairs). We find the total area that the staircases encompass, looking at the cross-section. Then, we multiply this by the width, $3$ , to get the volume. To find the area, we first note that the cross-section of the staircase is an infinite number of 45-45-90 right triangles, each with leg length $\tfrac {1}{n}$ .

Thus, the area of the $n$ th triangle is $\dfrac {\left( \dfrac {1}{n} \right)^2}{2}$ . Thus, the total area is

$\displaystyle\sum_{n=1}^{\infty} \dfrac {1}{2} \cdot \left( \dfrac {1}{n} \right)^2,$

so the volume is

$V = 3 \cdot \displaystyle\sum_{k=1}^{\infty} \dfrac {1}{2} \left( \dfrac {1}{n} \right)^2 = \dfrac {3}{2} \cdot \displaystyle\sum_{n=1}^{\infty} \dfrac {1}{n^2} = \dfrac {3}{2} \cdot \dfrac {\pi^2}{6} = \dfrac {\pi^2}{4},$

so our answer is $\left\lfloor 100 \cdot \dfrac {\pi^2}{4} \right\rfloor = \left\lfloor 25 \pi^2 \right\rfloor = \boxed {246}$ .

We note that the $n^{\textrm{th}}$ step has volume $3 \cdot \frac{1}{n} \cdot \frac{1}{2n} = \frac{3}{2n^2}$ . The volume of the staircase, then, is $\sum_{n=1}^{\infty} \frac{3}{2n^2} = \frac{3}{2} \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{4}$ . We then have $\left \lfloor{100N}\right \rfloor= \boxed{246}$

This is an idea of what the staircase looks like.

To find the area of this shape (not the volume yet; we will multiply by 3 later), we notice that the area of the $n^{\textrm{th}}$ step is the base times the height divided by two. We know both of these factors. Let $A_n$ denote the area of the $n^{\textrm{th}}$ step.

$A_n = \frac{1}{2} bh = \frac{1}{2} \left(\frac{1}{n}\frac{1}{n}\right) = \frac{1}{2} \left(\frac{1}{n^2}\right)$

So the volume of the entire staircase is the sum of all of these steps.

$\sum\limits_{i=1}^{\infty} A_i = \sum\limits_{i=1}^{\infty} \frac{1}{2} \left(\frac{1}{n^2}\right) = \frac{1}{2} \sum\limits_{i=1}^{\infty} \frac{1}{n^2}$

It is known that the infinite summation of the reciprocals of squares converges to $\frac{\pi^2}{6}$ . Also, remember that the stairs are not two-dimensional! They have a width of 3 that we will factor in right now.

$N = 3\sum\limits_{i=1}^{\infty} A_i = 3\frac{\pi^2}{12} = \frac{\pi^2}{4}$

$\lfloor 100N \rfloor = \lfloor 246.74011 \dots \rfloor = \boxed{246}$