Find the indicated sum n sigma k=1 4(0.5)^k

Answer 1

= 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