已知数列{an}中,其前n项和为sn,满足sn=2an-1,n∈N*,数列{bn}满足bn=1-log1/2an,n
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s1 = a1 = 2a1 -1
a1 = 1
sn - sn-1 = an = (2an-1) - (2an-1 -1) = 2an - 2an-1
an = 2an-1
通项an = 2^(n-1)
bn = 1-log1/2. an
= 1-log1/2. 2^(n-1)
= 1- log1/2. (1/2)^-(n-1)
= 1-(-(n-1))
= 1+(n-1)
=n
通项bn = n
a1 = 1
sn - sn-1 = an = (2an-1) - (2an-1 -1) = 2an - 2an-1
an = 2an-1
通项an = 2^(n-1)
bn = 1-log1/2. an
= 1-log1/2. 2^(n-1)
= 1- log1/2. (1/2)^-(n-1)
= 1-(-(n-1))
= 1+(n-1)
=n
通项bn = n
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