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#3 2011-04-12 13:21:01
- gAr
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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."
#4 2011-04-20 20:41:46
- alexa.pete9
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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.
#5 2011-04-20 20:46:59
- gAr
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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."
#6 2011-04-20 21:46:14
- bobbym
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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.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.
#7 2011-05-12 05:21:03
- reconsideryouranswer
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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 (2011-05-12 05:45:19)
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#8 2011-05-12 06:41:12
- reconsideryouranswer
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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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