Let's modify the Christmas Carol, "The Twelve Days of Christmas," for an algebraic reasoning task befitting the season:

How many gifts did your true love receive on each day? If the song was titled, "The Twenty-Five Days of Christmas," how many gifts would your true love receive on the twenty-fifth day? How many total gifts did she or he receive on the first two days? The first three days? The first four days? How many gifts did she or he receive on all twelve days? Can you create an equation, table, figure, graph to represent, "The X Days of Christmas," following the same pattern?

Who is up to this challenge in their classroom? If so, report back on how it played out. Let's have some fun with algebra!

source: twitter @ddanmeyer

## Comments

Just checking in - I see that 232 people viewed this post but no one tried it. Is it too difficult? Can we work it out together? What information do we need to consider to answer this question? Thoughts?

Day 1 - 1

Day 2 - 3

Day 3 - 6

Day 4 - 10

Day 25 - 324

Total gifts given over the twelve days - 364

Total gifts given over 25 days - 2925

The total for a day is the total for the day before plus the number of gifts given that day or the number of the day since both are the same.

Yes , I have used this lesson in class two weeks ago. We calculated the number of gifts per day and the number of gifts after each day to the total of 78 I believe on day 12 and 364 after 12 days. We also talked about tetrahedral numbers and drew the picture of gifts increasing in a Christmas tree shape. Then the students drew a line graph using the data to see the incremental increase. As follow up they were asked to calculate the number gifts for 25 days. We also wanted to create the model using magnetic balls, but didn't have time.

Thanks for the idea!

Did your students seem to like this lesson? What was the biggest take away from it? Just curious.

Brooke

Brooke has mentioned before that she would like to encourage more discussion of how the Common Core Mathematical Practices can be fostered in math classrooms. Among these practices are expectations that teachers will be pushing their students to generalize and look for connections to other parts of math. So I was hoping that when she suggested the simple context of counting and adding systematically the gifts on the 12 days of Christmas that it would lead to reports of some classroom discussion topics like the following. (I admit that as a retired math prof I don't teach math to (young) adults any longer, but I do every day tutor mostly ESL students (9-10 grade) in math at a local high school.)

We consider a sequence of days n: 1, 2, 3, 4, 5.... and on day n we are given n identical new gifts different from what we received on earlier days (along with the gifts from previous days). The context suggests that we consider the sequence S of partial sums given by S(n) = 1+2+3+...+n and then asks specifically for S(12) and S(25). For the next step, let T be the sequence of partial sums of the S sequence given by T(n) = S(1)+S(2)+S(3)+...+S(n). Then we are asked to compute T(12) and T(25). Of course we expect more than simply adding up the sequence terms from 1 to 12 or 25: that would be simple arithmetic, but this is supposed to be an introduction to an "Algebra Math Puzzle". We want to generalize -- find a systematic and efficient way to calculate S(n) and T(n) for any n.

The first part might remind us of the story of Gauss in the late1800's whose class in school was given the task of calculating S(100). It is said that he wrote down the correct answer very quickly. One way that he might have worked this out is the following.

2*S(100) = (1+2+... +99+100)+(100+99+...+2+1) = [1+100]+[2+99]+....+[99+2]+[100+1] = 100 *101 So S(100) = [100*101]/2.

Similarly, for any n we can easily see that S(n) = n*[n+1]/2. So S(12) = 78 and S(25) = 325.

But also notice that the S sequence is given by a quadratic polynomial in n. i.e. The partial sums of a "linear sequence are given by a quadratic sequence". Is that really true for ANY linear sequence? And if so, how can we find the correct quadratic polynomial that yields the partial sums?

For example, given Integers A < B, how can we calculate the sum of all the integers A, A+1, A+2, ..., A+n =B ? Notice that n = B - A but we want to add all n+1 of the numbers A+0, A+1, ...,B. Thus the sum is [n+1]*A + S(n) = [n+1]*A + n*[n+1]/2 --- again a quadratic function of n.

Consider the most general linear function G(n) = A + n*D and let M(n) = G(0)+G(1)+G(2)+...+G(n) then as above we can calculate M(n) = [n+1]*A +S(n)*D which is again a quadratic function of n.

Ok, so we should look a bit more closely at linear and quadratic functions and their corresponding sequences -- and it is usually most convenient to begin a sequence with the term number n=0. The constant function K(n) = 1 for all n, gives the sequence (1,1,1,1,1,1,1,...) and the identity function L(n) = n gives the sequence (0, 1, 2, 3, 4, 5, ... ) and the quadratic function Q(n) = n*[n-1]/2 gives the sequence (0, 0, 1, 3, 6 10, 15, ...). By 8th grade most of our students are familiar with calculating the terms of the first and second Forward Differences of a sequence; given a sequence W, the n-term of its first forward difference is V(n) = W(n+1) -- W(n) and the first forward difference U of V is then called the second difference of W. You can easily see that the first forward difference of Q above is L, and the first forward difference of L is K. (This is the discrete analogue of the derivative of [X^2]/2, X, and 1.) By 9th grade our students know that a sequence is quadratic if and only if its second difference is constant. Let's look at an example of this now.

n | 0 1 2 3 4 5 6 ....

W(n) | 5 1 0 2 7 15 26 ....

V(n) | -4 -1 2 5 8 11...

U(n) | 3 3 3 3 3 ....

But now it follows that W = 3*Q +(-4)*L +5*K, that is, W(n) = 3*n*[n-1]/2 +(-4)*n + 5 . In fact, we could use other columns in this table as coefficients if we shift Q, L, and K to the right. So we get lots more algebra exercises checking that we could also calculate W as follows:

W(n) = 3*[n-1]*[n-2]/2 +(-1)*[n-1] + 1 or as 3*[n-2]*[n-3]/2 + 2*[n-2} + 0 etc.

You might also come at this from another point of view -- the fact that all the binomial coefficients C(n, k) are available in most scientific calculators, or you can generate them easily in a spreadsheet where n is the row number (begin with 0), k is the column number (begin with 0), and Pascal's triangle identity C(n+1, k) = C(n, k) + C(n, k-1) generates the table recursively row by row beginning with row 0 = (1, 0, 0, 0, 0,..) and column 0 = (1, 1, 1,1,...). Then the special sequences K, L, Q above are columns 0, 1, and 2 of this spreadsheet, that is, K(n) =1 = C(n,0), L(n) = n = C(n, 1) and Q(n) = n*[n-1]/2 = C(n, 2). The first difference of any column of this spreadsheet is the previous column. Each entry C(n+1, k) in column k is the sum of all the earlier entries (0 to n) in the previous column. So now one should guess that the tetrahedral numbers T(n) in the second part of the 12 days of Christmas puzzle will be easily findable using columns 0 to 3 of the spreadsheet and will be computable from a cubic polynomial.

There are many other neat connections to the above that involve quadratics: Galileo and balls rolling down inclined planes, partial sum of consecutive odd numbers being a square (how does that show up when graphing parabolas in 9th grade?), the symmetric difference of a quadratic distance/time function giving instantaneous speed (without the derivative). And the geometry of parabolic reflectors (telescopes and flashlights).

Thank you Ladnor - does everyone understand this? It might be a complex so if you have questions about what Ladnor did - please ask!

Hi,

I don't really understand Ladnor's response and explanation, but I sort of follow it. Can't really ask questions, because it would take forever, but I hope in the future to be able to grasp the algebra better.