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Discussion: All in PreCalculus
Topic: series
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Subject:   RE: series
Author: Sione
Date: Mar 10 2004
On Mar 09, 2004, Wayne Hackney wrote:

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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