Video Transcript
A child wants to build a triangular wall using identical wooden cubes. The top row of the stack has one cube, the second row has two cubes, the third row has three cubes, and so on. Find the number of rows they can build with exactly 136 wooden cubes.
Let’s begin by sketching the wall this child is building. Now, obviously, the child is starting at the bottom of the wall. But once the wall is built, the top row has exactly one cube. The second row of the wall has two cubes. And then after this, the third row has three cubes. The wall continues until the child has used exactly 136 of these cubes.
We note that in each row, the child uses one more cube than they did in the previous row. The number of cubes in each row therefore forms an arithmetic sequence, because the terms increase by the same amount each time. The common difference, which we refer to as 𝑑, is one. And the first term, which we refer to as 𝑎 sub one, is also one, because this is the number of cubes in the top row. We’re asked to work out the number of rows the child can build when they use exactly 136 cubes. In terms of our sequence, this means we need to work out how many terms in our sequence have a sum equal to 136.
The formula for finding the sum of the first 𝑛 terms of an arithmetic sequence is 𝑆 sub 𝑛 is equal to 𝑛 over two multiplied by two 𝑎 sub one plus 𝑛 minus one multiplied by 𝑑. We know the values of 𝑎 sub one and 𝑑; they’re each one. And we know what we want 𝑆 sub 𝑛 to be. We want it to be 136. So we can form an equation. 136 is equal to 𝑛 over two multiplied by two times one plus 𝑛 minus one multiplied by one.
We now need to solve this equation to find the value of 𝑛. Multiplying both sides of the equation by two will eliminate the fraction on the right-hand side and give 272 on the left-hand side. At the same time, we simplify within the parentheses to give two plus 𝑛 minus one, which simplifies further to one plus 𝑛 or 𝑛 plus one. We can then distribute the parentheses on the right-hand side, giving 272 is equal to 𝑛 squared plus 𝑛. Finally, we’ll group the three terms on the same side of the equation, giving zero is equal to 𝑛 squared plus 𝑛 minus 272.
This is a quadratic equation in 𝑛. And it can in fact be solved by factoring. As the coefficient of 𝑛 squared is one, the first term in each set of parentheses is 𝑛. And we’re then looking for two numbers whose sum is the coefficient of 𝑛, that’s positive one, and whose product is the constant term, which is negative 272. Perhaps by listing the factors of 272 and using a bit of trial and error, we find that the two numbers which have these properties are positive 17 and negative 16. So the quadratic factors as 𝑛 plus 17 multiplied by 𝑛 minus 16 is equal to zero.
We then recall that if the product of two factors is equal to zero, then one of the factors themselves must be equal to zero. So either 𝑛 plus 17 equals zero, in which case 𝑛 equals negative 17, or 𝑛 minus 16 equals zero, in which case 𝑛 equals 16. Whilst these are both valid solutions to this quadratic equation, they aren’t both valid solutions to this problem. 𝑛 represents the number of rows the child can build. So it must be a positive integer. We therefore disregard the solution 𝑛 equals negative 17, and our answer is 𝑛 equals 16.
If we substitute this value of 𝑛 back into the formula for finding the sum of the first 𝑛 terms of an arithmetic sequence, we find that the sum of the first 16 terms is indeed equal to 136. So our answer is that if the child built a triangular wall in this way using exactly 136 cubes, they’ll be able to build 16 rows.
