Question

已知数列｛an｝是公比q>1的等比数列,且a1+a2=40,a1*a2=256,bn=log2an,求数列｛bn｝的通项

Answer 1

a1+a2=40,a1a2=256

a1,a2是方程x²-40x+256=0的两根.

(x-8)(x-32)=0

x=8或x=32

两根均为正,则a1,a2均为正,又公比q>1,因此a2>a1

a1=8 a2=32

q=a2/a1=32/8=4

an=a1q^(n-1)=8×4^(n-1)=2^(2n+1)

bn=log2(an)=log2[2^(2n+1)]=2n+1

数列{bn}的通项公式为bn=2n+1.