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

  • maths - ,

    = 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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