Posted by ELIJAH HAVA on Sunday, April 28, 2013 at 2:47am.
Find the indicated sum n sigma k=1 4(0.5)^k

maths  Reiny, Sunday, April 28, 2013 at 5:57am
= 4(.5)^1 + 4(.5)^2 + 4(.5)^3 + .... + 4(.5)^n
= 4( .5^1 + .5^2 + .5^3 + ...+ .5^n)
just for the brackets,
a = .5 , r = .5 , n = n
sum(n) = .5( 1  .5^n)/(1.5)
= 1  .5^n
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