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#1 2008-05-02 04:50:55

tony123
Member
Registered: 2007-08-03
Posts: 189

sum

Last edited by tony123 (2008-05-02 04:52:52)

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#2 2008-05-06 01:53:43

TheDude
Member
Registered: 2007-10-23
Posts: 361

Re: sum


Wrap it in bacon

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#3 2011-04-11 15:21:01

gAr
Member
Registered: 2011-01-09
Posts: 3,479

Re: sum

Hi,


"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense"  - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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#4 2011-04-19 22:41:46

alexa.pete9
Member
Registered: 2011-04-18
Posts: 11

Re: sum

Caushy Root Law is applicable on this question. And it can also be solved by applying the limits on it. The answer of the Question is 3.

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#5 2011-04-19 22:46:59

gAr
Member
Registered: 2011-01-09
Posts: 3,479

Re: sum

Can you show us how?
But the answer is 13.


"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense"  - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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#6 2011-04-19 23:46:14

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,481

Re: sum

Hi all;

I do not think there is much to show. You only have to sum the first 2 to see the sum is greater than 3.


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#7 2011-05-11 07:21:03

reconsideryouranswer
Member
Registered: 2011-05-11
Posts: 172

Re: sum

bobbym & reconsideryouranswer wrote:

Hi all;

I do not think there is much to show. You only have to sum the first 2 to see the  >> sum is greater than 3. <<

The above amendment comes right out of the first two consecutive terms, as well as
it shows 3 followed by an excess positive number.

This exemplifies the highlighted comment in the quote box.

Last edited by reconsideryouranswer (2011-05-11 07:45:19)


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#8 2011-05-11 08:41:12

reconsideryouranswer
Member
Registered: 2011-05-11
Posts: 172

Re: sum

Break up the original sum into the sum of these separate parts:


sum (1 to oo) [(n^2)/(2^n)]:


Let S = 1/2 + 4/4 + 9/8 + 16/16 + ...

Then (1/2)S = 1/4 + 4/8 + 9/16  + ...

S - (1/2)S = 1/2 + 3/4 + 5/8 + 7/16 + ...

(1/2)S = 1/2 + 3/4  + 5/8 + 7/16 + ...

(1/4)S = 1/4 + 3/8 + 5/16 + 7/32 + ...

(1/2)S - (1/4)S = 1/2 + 2/4 + 2/8 + 2/16 + ...

(1/4)S = 1/2 + 2(1/4 + 1/8 + 1/16 + 1/32 + ...)

(1/4)S = 1/2 + 2(1/2)

(1/4)S = 1/2 + 2/2

(1/4)S = 3/2

4(1/4)S = 4(3/2)

S = 6

---------------------------------------------------------

sum (1 to oo) [(2n)/(2^n)]:

= 2[sum (1 to oo) (n)/(2^n)]:

= 2[1/2 + 2/4 + 3/8 + 4/16 + ...] **

Let S = 1/2 + 2/4 + 3/8 + 4/16 + ...

(1/2)S = 1/4 + 2/8 + 3/16 + ...

S - (1/2)S = 1/2 + 1/4 + 1/8 + 1/16 + ...

(1/2)S = 1

2(1/2)S = 2(1)

S = 2


Then ** = 2[2]

= 4

---------------------------------------------------------

sum (1 to oo) [3/(2^n)]:

= 3[sum (1 to oo) 1/(2^n)]

= 3[1/2 + 1/4 + 1/8 + ...]

= 3[1]

= 3

--------------------------------------------------------

Then the sum equals

6 + 4 + 3 =

13


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