Question

数学试题 急！

已知数列{an}是等比数列，a1=1 a4=8.数列{bn}是等差数列，且b2=a2 b4=a3.求an,bn

Answer 1

解：{an}是等比数列，又a1=1 a4=8
a4=a1q^3
带入数据，解得q=2
所以 an=a1q^(n-1)=2^(n-1) 说明：2的n-1次方

b2=a2=a1q=1*2=2
b4=a3=a1*q^2=1*2^2=4
{bn}是等差数列 则有b4=b2+（4-2）d
解得d=1
所以bn=b1+（n-1）d
=b2-d+（n-1）d
=1+(n-1)1
=n

Answer 2

解：∵{an}是等比数列
∴a4=a1*q^3
4=1*q^3
q=4^(1/3)
∴an=1*q^(n-1)
=[4^(1/3)]^(n-1)
∴a2=a1*q
=1*4^(1/3)
=4^(1/3)
a3=a1*q^2
=2*2^(1/3)
∴b2=a2 =4^(1/3)=2^(2/3)
b4=a3= 2*2^(1/3)=2^(4/3)
∴d=1/2*(b4-b2)
=2^(-1)*2^(4/3)-2^(-1)*2^(2/3)
=2^(1/3)-2^(-1/3)
∴b1=b2-d
=2^(2/3)-2^(1/3)+2^(-1/3)
∴bn=b1+(n-1)d
=2^(2/3)-2^(1/3)+2^(-1/3)+(n-1)[2^(1/3)-2^(-1/3)]