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Topic:  series 
Related Item:  http://mathforum.org/mathtools/tool/633/ 
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Subject:  RE: series 
Author:  Sione 
Date:  Mar 10 2004 
On a recent test there was a symbol that looked like pi.
It wanted the sum of a series. What operation do you do to get the sum?
8 n
II {1/2)
n=0
The pi symbol means products not sums.
=({1/2}^0)*({1/2}^1)*({1/2}^2)*({1/2}^3)*({1/2}^4)*({1/2}^5)*({1/2}^6)*({1/2}^7)*({1/2}^8)
= (1)*(1/2)*(1/4)*(1/8)*(1/16)*(1/32)*(1/64)*(1/128)*(1/256)
= 1/(2^36)
To sum it up:
n=8 n
II {1/2} = {(1)^n} / { 2^(1/2*n*(n+1)) }
n=0
Say n=8 , then:
= {(1)^n} / { 2^(1/2*n*(n+1)) }
= {(1)^8} / { 2^(1/2*8*(8+1)) }
= { 1 } / { 2^( 36 ) }
= 1 / 2^( 36 )
Cheers,
Sione.
 
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