已知数列{an}是公差为2的等差数列,其前n项和为sn,数列{bn}是首项2的等比数列,且b3s2=32.b1b3=b4求数...

2025-05-14 12:41:33
推荐回答(2个)
回答1:

您好!

b1b3=b4
b1*b1q^2=b1q^3
b1q^2(b1-q)=0
b1=2,q≠0,
故q=b1=2
bn=2^n
b3=8
s2=32/b3=4=a1+a2
a2-a1=d=2
故a1=1,a2=3
an=2n-1

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回答2:

an=a1+(n-1)d
=a1+2(n-1)
bn=b1*q^(n-1)
=b1*2^(n-1)
b3s2=32即b1*2^2*(a1+a1+2)=32
b1b3=b4即b1*b1*2^2=b1*2^3
解得:a1=1
b1=2
an=1+2(n-1)
=2n-1
bn =2*2^(n-1)
=2^n