>>>>>Looks like the Summation of k for k=1 to n. Solution to this is
>>>>>
>>>>>( n * ( n + 1 ) ) / 2
>>>>
>>>>Er, not exactly. That gets you the number of items on the nth day. What we're seeking is the sum as described above, where each day is cumulative from previous days. Day 3, you get 6+3+1=10 items for example, all the items from days 1, 2 , and 3...(Like the twelve days of Xmas) :~)
>>>
>>>Bill's solution works. On day 3, n=3 so [3 * ( 3 + 1)] / 2 = 6 which is 3+2+1. Day 4, (4 * 5)/2 = 10, or 4+3+2+1 = 10.
>>
>>but total gifts for day 4 = 10 + 6 + 4 + 1
>
>summation (i = 1 to n) of (i squared + i)/2...

Er, sorry, Dave, but there is no summation (looping) allowed in the solution, it's only a one-liner...