Posers and Puzzles
Sum the series:
1*2*3-2^3+3*4*5-4^3+...+9999*10000*10001-10000^3.
- Joined
- 12 Mar '03
- Moves
- 44411
- 25 005 000
edit: but of course, it is not real 😀
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'. 🙂
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)
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
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