### 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.So how many presents did I receive all together?

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.

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,

*δ*is the first difference (the difference between successive terms of

^{1}*P(n)*— which is

*p(n)*of course),

*δ*is the second difference (that between successive terms of

^{2}*δ*) and

^{1}*δ*is the third difference.

^{3}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 | ... |

δ^{1} | 1 | 3 | 6 | 10 | 15 | 21 | ... | |

δ^{2} | 2 | 3 | 4 | 5 | 6 | ... | ||

δ^{3} | 1 | 1 | 1 | 1 | ... |

As the

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

^{3}[1]

*P(n)*= a

*n*

^{3}+ b

*n*

^{2}+ c

*n*+ 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}⁄_{6}n^{3} + ^{1}⁄_{2}n^{2} + ^{1}⁄_{3}n |

= ^{1}⁄_{6}n(n^{2} + 3n + 2) | |

= ^{1}⁄_{6}n(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.

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