Twelve Days of Christmas

This is a classic problem. You can show that this general series can be solved by the equation $\frac{1}{6}(n)(n+1)(n+2).$

Actually I think the Question is wrong it should be how many gifts in total does the true lover sends to his beloved . If it is for how many gifts does the true lover sends by the twelfth day? The solution will obviously be

= 1+2+3+4+5+6+7+8+9+10+11+12

=n/2*(a+l) =78

for the total number of gifts it should be

=1 * 12+2 * 11+3 * 10+4 * 9+5 * 8+6 * 7+7 * 6+8 * 5+9 * 4+10 * 3+11 * 2+12 * 1

=(1 * 12+2 * 11+3 * 10+4 * 9+5 * 8+6 * 7) * 2

=364

This has been written considering that he sends he sends 1 gift on the first on the first day , 1+2 gifts on the second day and so on.

On the first day, there is 1 gift. On the second, there is 3, then 6, then 10. These are the triangular numbers.

$0 + 1 = 1$ $1 + 2 = 3$ $3 + 3 = 6$ $6 + 4 = 10$ $10 + 5 = 15$

First, there is 0, then keep adding 1, 2, 3, 4, 5, so on, and you will get the triangular numbers. The sum of all of the gifts is obviously the sum of the first 12 triangular numbers, namely, 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 + 66 + 78. The triangular numbers can also be expressed in groups of objects that form triangles.