Question

已知等比数列{an}的公比为q,若a[(n+1)/2]=m(n为奇数),则a[(3n+1)/2]=

Answer 1

a[(n+1)/2]=a1q^[(n+1)/2 -1]=a1q^[(n-1)/2]=m
a1=m/q^[(n-1)/2]=m×q^[(1-n)/2]
a[(3n+1)/2]=a1q^[(3n+1)/2 -1]
=a1q^[(3n-1)/2]
=m×q^[(1-n)/2]×q^[(3n-1)/2]
=m×q^[(1-n)/2 +(3n-1)/2]
=m×q^n

Answer 2

a(n) = a(1)*q^(n-1)

a[(n+1)/2]
= a(1)*q^[(n+1)/2-1]
= a(1)*q^[(n-1)/2]
= m
a(1) = m/q^[(n-1)/2]

a[(3n+1)/2]
= a(1)*q^[(3n+1)/2-1]
= m/q^[(n-1)/2] * q^[(3n+1)/2-1]
= m/q^[(n-1)/2] * q^[(3n-1)/2]
= m*q^[(3n-1)/2 - (n-1)/2]
= m*q^[(2n)/2]
=mq^n