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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