Question

高一等比数列证明题,求助!

正数列{an}和{bn}满足,对于任意自然数n,an,bn,a(n+1)成等差数列,bn,a(n+1),b(n+1)成等比数列. 证明:数列{根号bn}为等差数列

Answer 1

[a(n+1)]^2=bnb(n+1)
a(n+1)=√[bn*b(n+1)]
2bn=an+a(n+1)
2bn=√[bn*b(n-1)]+√[bn*b(n+1)]
2√bn=√b(n-1)+√b(n+1)
所以数列{√bn}为等差数列

Answer 2

a(n+1)=√[bn*b(n+1)]

2bn=an+an+1

2bn=√[bn*b(n-1)]+√[bn*b(n+1)]

2√bn=√b(n-1)+√b(n+1)

所以数列{√bn}为等差数列