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Vert
@Vert
January 2020
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Determine o 15° terno da PG (25 6,128,64,31...)
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Karinapontes
PG u1 = 256 u2 = 128 razão q = u2/u1 = 128/256 = 1/2 un = u1*q^(n - 1) u15 = 256*(1/2)^14 u15 = 2^8/2^14 = 2^-6 = 1/64 pronto
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Eduardo2018
(256,128,64,32...)
a1=256
an=?
q=1/2
n=15
An=a1,q^n-1
An=256.(1/2)^15-1
An=256.(1/2)^14
An=256.(1/2)
An=256/16.384
An=256:256/16.384:256
An=1/64
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Lista de comentários
a1=256
an=?
q=1/2
n=15
An=a1,q^n-1
An=256.(1/2)^15-1
An=256.(1/2)^14
An=256.(1/2)
An=256/16.384
An=256:256/16.384:256
An=1/64