How Many Presents from my True Love?

On the first day of Christmas my true love gave to me a partridge in a pear tree.
On the second day of Christmas my true love gave to me two turtle doves and a partridge in a pear tree.
...
On the twelfth day of Christmas my true love gave to me twelve drummers drumming, eleven pipers piping, ten lords-a-leaping, nine ladies dancing, eight maids-a-milking, seven swans-a-swimming, six geese-a-laying, five golden rings, four calling birds, three french hens, two turtle doves and a partridge in a pear tree.

So how many presents did I receive all together?

This can be solved using the differences method, which is nicely explained by Ken Ward.

In the table below, n represents the day of Christmas (starting from 0, the day before I start receiving presents), p(n) represents the number of presents I receive on that day, P(n) represents the total number of presents received so far, δ¹ is the first difference (the difference between successive terms of P(n) — which is p(n) of course), δ² is the second difference (that between successive terms of δ¹) and δ³ is the third difference.

n	0	1	2	3	4	5	6	...
p(n)	0	1	3	6	10	15	21	...
P(n)	0	1	4	10	20	35	46	...
δ1		1	3	6	10	15	21	...
δ2			2	3	4	5	6	...
δ3				1	1	1	1	...

As the δ³ values are all the same, the equation we're after is a cubic polynomial.  That is, it's of the form:

[1]P(n) = an³ + bn² + cn + d

Now we can do some substitution.  We can see immediately that

P(0) = 0a + 0b + 0c + d = 0 ⇒ d = 0

And now, from the next three terms, we can create 3 simultaneous equations:

[2]P(1) = a + b + c = 1
[3]P(2) = 8a + 4b + 2c = 4
[4]P(3) = 27a + 9b + 3c = 10

We can eliminate c, first from [2] and [3] and then from [2] and [4] to produce:

[5]6a + 2b = 2
[6]24a + 6b = 7

From which we can deduce that (24 – 18a) = 1 ⇒ 6a = 1, or a = ¹⁄₆.

Substituting that back in [5], we find that 1 + 2b = 2 ⇒ 2b = 1, or b = ¹⁄₂.

Finally we can use [2] to determine that ¹⁄₆ + ¹⁄₂ + c = 1, or that c = ¹⁄₃.

Now we can substitute these values back in [1] and tidy up a bit:

P(n)	= 1⁄6n3 + 1⁄2n2 + 1⁄3n
	= 1⁄6n(n2 + 3n + 2)
	= 1⁄6n(n + 1)(n + 2)

So, by the 12^th day of Christmas I will have received ¹⁄₆.12(12 + 1)(12 + 2) or 2.13.14 = 364 presents. Not bad going.