Question

已知数列{an}和{bn}满足a1a2a3...an=（根号2）^bn.若{an}为等比数列,且a1=2,b3=6+b2

Answer 1

a1...an=√2^bn

a1...ana(n+1)=√2^b(n+1)

两式相除得：a(n+1)=√2^[b(n+1)-bn]

由此即可看得an>0

且a3=√2^(b3-b2)

即a3=√2^6=8

故a3/a1=8/2=4=q²，得q=2

an=2^n

从而a1...an=2^(1+2+...+n)=2^[n(n+1)/2]=√2^n(n+1)

对比已知条件，得bn=n(n+1)