已知数列{an}满足an-1 -2an +an+1=0 且a1=2 a3=4 数列{bn}的前n项和为Sn=2bn-1

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已知数列{an}满足an-1 -2an +an+1=0 且a1=2 a3=4 数列{bn}的前n项和为Sn=2bn-1
已知数列{an}满足an-1 -2an +an+1=0 且a1=2 a3=4 数列{bn}的前n项和为Sn=2bn-1

已知数列{an}满足an-1 -2an +an+1=0 且a1=2 a3=4 数列{bn}的前n项和为Sn=2bn-1
(1)
a(n-1) -2an +a(n+1)=0
a(n+1)-an = an - a(n-1)
=a3-a2 (1)
a(n+1)- an = a2 - a1 (2)
(1)+(2)
2[ a(n+1) -an ] = a3 +a1
=6
an -a(n-1) =3
an - a1 = 3(n-1)
an = 3n-1
Sn = 2bn-1
n=1 ,b1=1
for n>=2
bn = Sn -S(n-1)
=2bn - 2b(n-1)
bn =2b(n-1)
=2^(n-1) .b1
=2^(n-1)
(2)
cn =[log(an -1) ]
=[log(3n)]
c1= [log3] =1
c2= [log6] =2
T2 = c1+c2 =1+2 = 3