已知等差数列{an}满足a2=3,a5=9,若数列{bn}满足b1=3,bn-1=a下标bn则bn为?
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已知等差数列{an}满足a2=3,a5=9,若数列{bn}满足b1=3,bn-1=a下标bn则bn为?
n-1里的n-1也是下标吧?
d=(a5-a2)/3=2
an=a2+(n-2)d=2n-1
∴a(bn)=2bn - 1
又∵b(n-1)=a(bn)
∴2bn - 1=b(n-1)
2(bn - 1)=b(n-1) - 1
∴bn - 1=[b(n-1) - 1]/2
∴数列{bn - 1}是以b1 - 1=2为首项,1/2为公比的等比数列.
(注:这里不用讨论n=1时了,因为根据上式已经有b2 - 1=(b1 -1)/2了)
∴bn - 1=2×(1/2)^(n-1)=1/2^(n-2)
∴bn=1/2^(n-2) + 1
d=(a5-a2)/3=2
an=a2+(n-2)d=2n-1
∴a(bn)=2bn - 1
又∵b(n-1)=a(bn)
∴2bn - 1=b(n-1)
2(bn - 1)=b(n-1) - 1
∴bn - 1=[b(n-1) - 1]/2
∴数列{bn - 1}是以b1 - 1=2为首项,1/2为公比的等比数列.
(注:这里不用讨论n=1时了,因为根据上式已经有b2 - 1=(b1 -1)/2了)
∴bn - 1=2×(1/2)^(n-1)=1/2^(n-2)
∴bn=1/2^(n-2) + 1
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