What is the equation for the 12 Days of Christmas?

Question

I need to find an equation that will help me figure how many gifts in total was recive in the 12 Days of Christmas. The equaion itself should also help figure out 20 Days of Christmas.

Answer 1

Start with 1 day:
f(1) = 1 partridge = 1 present

Then two days:
f(2) = 1 partridge + 2 turtle doves = 3 presents

Day 3:
f(3) = 1 + 2 + 3 = 6 presents

Day 4:
f(4) = 1 + 2 + 3 + 4 = 10 presents.

The sum of consecutive digits can be computed as:
f(n) = n(n+1) / 2

But I think what you are asking for is the *total* sum of presents.

S(1) = 1 = 1
S(2) = 1 + 3 = 4
S(3) = 1 + 3 + 6 = 10
S(4) = 1 + 3 + 6 + 10 = 20
S(5) = 1 + 3 + 6 + 10 + 15 = 35

Can you figure the formula for S(n)?

You can check the Online Encyclopedia of Integer Sequences, if you don't know the answer...

The sum of triangle numbers is the pyramidal (or tetrahedral) numbers given by the following formula.

S(n) = n(n+1)(n+2)/6

For S(12) = (12x13x14) / 6 = 364 gifts
For S(20) = (20x21x22) / 6 = 1,540 gifts

Incidentally, if you are familiar with binomials, the results are equivalent to C(n+2, 3).
(12 + 2) choose 3 = 364
(20 + 2) choose 3 = 1,540.

Answer 2

For the each and on a daily basis volume despatched right here Algebraic formula works: x^2/2 + x/2 the place x is the day style. occasion, say day 6, inputting 6 into the formula supplies 36/2 + 6/2 = 18 + 3 =21. For the cumulative finished the Algebraic formula works: x^3/6 + x^2/2 + x/3 the place x is the day style. occasion, say day 12, inputting 12 into the formula supplies: 288 + seventy two + 4 = 364

Answer 3

Bah, I should have known this was a well-known sequence.

Renny B's second answer is correct for the total number of gifts received.

Humbug.

--------------

Neat!

You receive 12 partridges in a pair tree. 12 x 1
11 sets of 2 turtle doves. 11 x 2
10 sets of 3 french hens. 10 x 3
9 sets of 4 calling birds. 9 x 4
....
1 set of 12 drummers drumming 1 x 12

So one way to calculate it for n days of Christmas would be to calculate:

If n is even:
2[(1)(n) + (2)(n-1) + (3)(n-2)...(n/2-1)(n/2)]

If n is odd:
2[(1)(n) + (2)(n-1) + (3)(n-2)...(floor(n/2))(ceil(n/2) + 1)] + ceil(n/2)^2

A little messier than I expected but I think the formulas do the trick.

Working on another way to think of this....

Answer 4

On the nth day of Christmas, my true love does indeed send to me a triangular number of gifts.

But if you want to know the today number of gifts received up to and including the nth day, then you need to sum these triangular numbers using the formula
n(n+1)(n+2)/6

Answer 5

What you're looking for are TRIANGULAR NUMBERS. They are the sum of 1, 2, 3, 4, ..., up to a number N.

The formula for the N-th triangular number is:
(n-squared + n) / 2