已知数列{an}是等差数列,a2=6,a5=18;数列{bn}的前n项和是Tn,且Tn+(1/2)bn=1.(1)求证：数列{bn}是等比数列(2)记cn=ab*bn,求{cn}的前n项和Sn
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已知数列{an}是等差数列,a2=6,a5=18;数列{bn}的前n项和是Tn,且Tn+(1/2)bn=1.(1)求证：数列{bn}是等比数列(2)记cn=ab*bn,求{cn}的前n项和Sn
已知数列{an}是等差数列,a2=6,a5=18;数列{bn}的前n项和是Tn,且Tn+(1/2)bn=1.
(1)求证：数列{bn}是等比数列
(2)记cn=ab*bn,求{cn}的前n项和Sn
已知数列{an}是等差数列,a2=6,a5=18;数列{bn}的前n项和是Tn,且Tn+(1/2)bn=1.(1)求证：数列{bn}是等比数列(2)记cn=ab*bn,求{cn}的前n项和Sn
a5-a2=3d=18-6
d=4
a1=a2-d=2
an=4n-2
Tn+1/2bn=1
所以T(n-1)+1/2b(n-1)=1
相减,Tn-T(n-1)=bn
bn+1/2bn-1/2b(n-1)=0
3/2bn=1/2n(n-1)
bn/b(n-1)=1/3
所以是等比数列
T1=b1
所以b1+1/2b1=1
b1=2/3,q=bn/b(n-1)=1/3
bn=2/3*(1/3)^(n-1)=2*(1/3)^n
cn=(4n-2)*2*(1/3)^n=4*(2n-1)*(1/3)^n
令x=Sn/4=1*(1/3)+3*(1/3)^2+……+(2n-1)*(1/3)^n
(1/3)x=1*(1/3)^2+3*(1/3)^3+……+(2n-3)*(1/3)^n+(2n-1)*(1/3)^(n+1)
(1/3)x-x=(2n-1)*(1/3)^(n+1)-2*[(1/3)^n+……+(1/3)^2]-1/3
=(2n-1)*(1/3)^(n+1)-2*(1/9)[1-(1/3)^(n-1)]/(1-1/3)-1/3
=(2n-1)*(1/3)^(n+1)-2/3+(1/3)^n
x=(3/2)(2n-1)*(1/3)^(n+1)-1+(3/2)(1/3)^n
=(n+1)*(1/3)^n-1
所以Tn=4x=4(n+1)*(1/3)^n-4