 Posers and Puzzles

1. 22 May '05 06:42
Sum the series:

1*2*3-2^3+3*4*5-4^3+...+9999*10000*10001-10000^3.
2. 22 May '05 10:19
-250500
3. 22 May '05 10:451 edit
- 25 005 000

edit: but of course, it is not real 😀
4. 22 May '05 11:071 edit
Originally posted by Mephisto2
- 25 005 000

edit: but of course, it is not real 😀
Yes, I was going to add '...provided negative numbers exist'. 🙂
5. 17 Jun '05 11:53
Originally posted by Mephisto2
- 25 005 000

edit: but of course, it is not real 😀
I was wondering if someone smarter than me can explain how they got an answer.

All i have is;

If I split the entire series into two parts

(1*2*3+3*4*5+.........)-(2^3+4^3+.......)

The first part can be written as Summation[n(n+1)(n+2)]

Summation (n^3 + 3(n^2) + 2n)

For the second part take 2^3 common
2^3(1+2^3 + 3^3......)
2^3 Summation(n^3)
6. 17 Jun '05 12:101 edit
Oops, looks like I summed up to 500 rather than 5000.

The general term is
(2n-1)(2n)(2n+1) - (2n)^3
= -2n

Summing between 1 and 5000 gives -25005000