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Lost Maths Worker science forum beginner
Joined: 03 May 2005
Posts: 18
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Posted: Fri Jun 17, 2005 11:59 am Post subject:
Sum the series
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1*2*3-2^3+3*4*5-4^3+...+9999*10000*10001-10000^3.
All i have is, any help as always is appreciated;
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) |
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Torsten Hennig science forum Guru Wannabe
Joined: 28 Apr 2005
Posts: 136
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Posted: Fri Jun 17, 2005 1:03 pm Post subject:
Re: Sum the series
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| Quote: | 1*2*3-2^3+3*4*5-4^3+...+9999*10000*10001-10000^3.
All i have is, any help as always is appreciated;
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)
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Hi,
(2i-1)*2i*(2i+1) - (2i)^3 = -2i.
So your sum equals
sum_i=1^5000 (-2*i) = -2*5000*5001/2 = -5000*5001.
Best wishes
Torsten. |
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Lost Maths Worker science forum beginner
Joined: 03 May 2005
Posts: 18
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Posted: Fri Jun 17, 2005 2:00 pm Post subject:
Re: Sum the series
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Thanks, i was also looking for a more easier solution, or more elegant, but now i wonder if there is, any help? |
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Julio Guerra science forum beginner
Joined: 18 Jun 2005
Posts: 1
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Posted: Sat Jun 18, 2005 4:37 pm Post subject:
Re: Sum the series
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I don't agree with you, Torsten's sollution is elegant and simple.
You have a mistake in grouping the first part to n(n+1)(n+2) and that's why
you couldn't solve it.
I was following your first steps and couldn't solve either.
Congratulations Torsten.
Regards
"Lost Maths Worker" <gpaohi@hotmail.com> wrote in message
news:13295445.1119016856703.JavaMail.jakarta@nitrogen.mathforum.org...
| Quote: | Thanks, i was also looking for a more easier solution, or more elegant,
but now i wonder if there is, any help? |
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