The Gifts of Christmas

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 :

\(\begin{array}{c | cc}
\text{Day}& \text{Count} & \text {Total}\\
\hline
1 & 1 & 1\\
2 & 1 + 2 & 3\\
3 & 1 + 2 + 3 & 6\\
4 & 1 + 2 + 3 + 4 & 10\\
n & 1 + 2 + 3 + 4 +… +n & \frac {n (n + 1) }{2} \end{array}\)

OldTrout, Going Postal

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) :

\( \begin {array}{c |c cc}
n & \text{Square} &  \text{Count} & \text{Total}\\
\hline
1 & 1 & 1 & 1\\
2 &4 & 1 + 4 & 5\\
3 & 9 & 1 + 4 + 9 & 14\\
4 & 16 & 1 + 4 + 9 + 16 & 30\\
n &  n^2 & 1 + 4 + 9 + 16 + … +n^2 & \frac { n(n + 1)(2n +1)}{6}
\end {array} \)

OldTrout, Going Postal

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 :

\(\begin{array}{c |c c c}
\text{Day} & \text{Triangle} &\text{Count} & \text {Total}\\
\hline
1 &1 & 1 & 1\\
2 &3 & 1 + 3 & 4\\
3 &6 & 1 + 3 + 6 & 10\\
4 &10 & 1 + 3 + 6 + 10 & 20\\
n &\frac{n(n +1)}{2} & 1 + 3 + 6 + 10 + … +\frac{n(n + 1)}{2} & \frac{n(n +1)(n +2)}{6} \end{array} \)

OldTrout, Going Postal

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