Lucas builds staircases from cubes. Each step is two cubes wide. He records his observations in a table:
<table>
<tr><td>Staircase height</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td></tr>
<tr><td>Cubes in the bottom row</td><td>\(2\)</td><td>\(4\)</td><td>\(6\)</td></tr>
<tr><td>Total number of cubes</td><td>\(2\)</td><td>\(6\)</td><td>\(12\)</td></tr>
</table>
a) For a staircase with height \(10\), how many cubes are in the bottom row, and how many cubes are needed altogether?
b) Let \(h\) be the staircase height, \(b\) the number of cubes in the bottom row, and \(t\) the total number of cubes. Write equations that express \(b\) and \(t\) in terms of \(h\).
Hints
- Compare the staircase height with the bottom-row count in each table column.
- For the total, examine how the accumulated number of cubes changes as the height increases.
- In part b), write each changing cube count as a dependent quantity determined by \(h\).
Solution
a) The bottom row has twice as many cubes as the height, so \(b = 2 \times 10 = 20\). The row counts are \(2, 4, 6, \ldots, 20\), whose total is \(110\) cubes.
b) The bottom-row relationship is \(b = 2h\). The total number of cubes is the sum of the first \(h\) even numbers, which is \(t = h(h + 1)\). For \(h = 10\), this gives \(t = 10 \times 11 = 110\).
Answer
a) Bottom row: \(20\) cubes; total: \(110\) cubes
b) \(b = 2h\) and \(t = h(h + 1)\)