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Gjeni shumen e 20 kufizave te fillimit te progresionit aritmetik √12,√3

 Jan 5, 2016
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This is a geometric sequence: sqrt(12), sqrt(3), sqrt(3/4), sqrt(3/16), sqrt(3/64), sqrt(3/256).....and so on. Keep multiplying the denominator by 4, so:256 X 4=3/1024, 3/4096.........etc.

 

I'm NOT sure as to you want: The sum of the 1st. 20 terms is:6.9281966............etc.

The 20th term is 3/68,719,476,736.

 

P.S. This is in Albanian language. 

 Jan 5, 2016

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