If you sent all the gifts in the Twelve Days of Christmas, then how many would you send?

The gifts for each day are the triangular numbers :

\(\begin{array}{c | cc}

\text{Day}& \text{Count} & \text {Total}\\

\hline

1 & 1 & 1\\

2 & 1 + 2 & 3\\

3 & 1 + 2 + 3 & 6\\

4 & 1 + 2 + 3 + 4 & 10\\

n & 1 + 2 + 3 + 4 +… +n & \frac {n (n + 1) }{2} \end{array}\)

We have done these before :

\( \displaystyle \sum_{k=1}^n k = \frac {n (n + 1)}{2} \)

In English, the sum from 1 to \(n\) of the counting numbers is equal to the last number multiplied by the last number plus one all divided by two.

On the twelfth day you would send :

\(\displaystyle \frac {12 \cdot 13}{2} = 6 \cdot 13 = 78 \) gifts.

This does not answer the question. How many gifts would be sent in total over the period of twelve days?

We have to sum the triangular numbers; that is, what is the sum of ?

\( \displaystyle \sum_{k=1}^n \frac {k(k+1)}{2} \)

This is equivalent to (take out the factor and expand)

\( \displaystyle \frac 12 \sum_{k=1}^n k^2 + k \)

Now, we know the sum of \(k\), so what is the sum of \(k^2\)? Don’t forget that the sum of the odd numbers \((2k + 1, k = 0,1,2,…)\) is \(n^2\). We have also done this before. Pay attention.

When we sum the square numbers we arrive at the square pyramid numbers (cannonballs) :

\( \begin {array}{c |c cc}

n & \text{Square} & \text{Count} & \text{Total}\\

\hline

1 & 1 & 1 & 1\\

2 &4 & 1 + 4 & 5\\

3 & 9 & 1 + 4 + 9 & 14\\

4 & 16 & 1 + 4 + 9 + 16 & 30\\

n & n^2 & 1 + 4 + 9 + 16 + … +n^2 & \frac { n(n + 1)(2n +1)}{6}

\end {array} \)

Now that we have all the information we can say that the sum of the triangular numbers is

\( \displaystyle \begin{align}\sum_{k=1}^n \frac{k(k+1)}{2} & = \frac 12 \Bigl (\frac {n(n+1)(2n+1)}{6} + \frac {n(n+1)}{2} \Bigr ) \\ & =\frac {n(n + 1)(n +2)}{6} \end{align}\)

These are the tetrahedral (triangular pyramid) numbers :

\(\begin{array}{c |c c c}

\text{Day} & \text{Triangle} &\text{Count} & \text {Total}\\

\hline

1 &1 & 1 & 1\\

2 &3 & 1 + 3 & 4\\

3 &6 & 1 + 3 + 6 & 10\\

4 &10 & 1 + 3 + 6 + 10 & 20\\

n &\frac{n(n +1)}{2} & 1 + 3 + 6 + 10 + … +\frac{n(n + 1)}{2} & \frac{n(n +1)(n +2)}{6} \end{array} \)

Hence, if you sent all the gifts in the Twelve Day of Christmas, then you would send :

\(\displaystyle \frac{12\cdot13\cdot14}{6} = 2\cdot13\cdot14 = 364 \) gifts.

© OldTrout \(2018\)

**No Audio file – it does not translate well**