In this solution we count the cubes in rows (or layers):
- The top layer has a single cube. The layer below has one cube beneath the top cube plus \(4\) new ones making a total of \(5\). The third layer below has cubes below these \(5\) plus \(4\) new ones to make \(9\). Continuing to add four each time gives a total of $$ 1+5+9+13+17+21 = 66 $$ cubes in the skeleton tower with six layers.
- Building upon the reasoning established in a), the number of cubes in the bottom (\(12\)th) layer will be \(1 + 4 \times 11\) since it is \(11\) layers below the top. So for this total, we need to add $$ 1 + 5 + 9+ \ldots + 45 .$$ One way to do this would be to simply add the numbers. A quicker method which will work for any number of layers is the so-called Gauss method (which the great mathematician reportedly used to sum a large list of whole numbers when he was very young). The idea is to rewrite the sum backward as $$45 + 41 + 37 + \ldots + 1.$$ Now if this list is placed below the previous list we can see that each column sums to \(46\). There are \(12\) columns so the answer to this problem is half of \(12 \times 46 = 552\) or \(276\).
- Let \(f(n)\) be the number of cubes in the \(n^{\rm th}\) layer counting down from the top. Then \(f(1)=1\) is the number of cubes in the top layer, \(f(2)=5\) the number of cubes in the second layer from the top and so on. In this case, the values of the function \(f(1), f(2), f(3), \ldots\) form an arithmetic sequence, where each term is obtained from the previous one by adding \(4\). In general, we have \(f(n) = 4(n-1) + 1\). The method used to solve problem b) provides a method both to solve the skeleton tower problem and to add a general arithmetic sequence. For the skeleton tower problem, with \(n\) layers in the tower, the sum giving the total number of cubes will be $$ 1 + 5 + 9 + \cdots + f(n) = 1 + 5 + 9 + \cdots + (4(n-1)+1). $$ If we use the method of problem b) here, twice this sum will be equal to \(n(4(n-1)+2)\) and so the general solution for the number of cubes in a skeleton tower with \(n\) layers is $$ \frac{n(4(n-1)+2)}{2} = n(2n-1). $$
Commentary
This problem is a quadratic function example. The other tasks in this set illustrate F.BF.1a in the context of linear (Kimi and Jordan), exponential (Rumors), and rational (Summer Intern) functions.
This is based on a task developed by the MARS/ Shell Centre team Mathematics Assessment Resource Service(adapted with permission for redistribution with the Creative Commons License in this form).