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

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

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

Wrap it in bacon

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

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

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

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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,237

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

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

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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