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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

Wrap it in bacon

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

gAr
Member
Registered: 2011-01-09
Posts: 3,481

### 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: 3,481

### Re: sum

Can you show us how?

"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
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### 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.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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

Member
Registered: 2011-05-11
Posts: 171

### 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. <<

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

Member
Registered: 2011-05-11
Posts: 171

### 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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