Sum the series:

1*2*3-2^3+3*4*5-4^3+...+9999*10000*10001-10000^3.

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- May 21st 2005, 11:37 PMcornothSum the series
Sum the series:

1*2*3-2^3+3*4*5-4^3+...+9999*10000*10001-10000^3. - May 22nd 2005, 12:11 AMaskmemath
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)

I hope you can expand for the individual summations.

For the second part take 2^3 common

2^3(1+2^3 + 3^3......)

2^3 Summation(n^3)

Hope you can take it from here. - May 23rd 2005, 02:19 AMcornoth
- 25,005,000 Correct?

Each 'group' starts from -2, then add -3+....+-9999+-10000 - Aug 24th 2005, 08:53 PMardnas
Can someone explain why the summation is like that? I still don't get it.

- Aug 25th 2005, 06:52 AMaskmemath
Can you please explain which part of the summation you didnt understand??

- Aug 25th 2005, 06:57 AMardnas
*For the second part take 2^3 common*

2^3(1+2^3 + 3^3......)

2^3 Summation(n^3)

This please.

Then do I add the two summation together? And should I put 5000 for n? - Aug 25th 2005, 07:24 AMaskmemath
1^3 + 2^3 + 3^3 + 4^3 +........5000^3

Each number can be written as n^3.

Since we are adding the numbers we take Summation(n^3)

Correct n=5000 since we took 2 common.

There's a negative sign between the two series so techincally you are subtracting one from the other. - Aug 25th 2005, 03:32 PMardnas
I understand what you are explaining, but how do I calculate or show my working for the whole thing?