Question

已知各项均为正数的两个数列{an}和{bn}满足：a(n+1)=(an+bn)/√(an²+bn²),n∈N+

设b(n+1)=(√2)bn/an,且{an}是等比数列,求a1和b1的值

Answer 1

{an},{bn}均为正项数列
{an}是等比数列
an=q^(n-1)*a1
a(n+1)=q^n*a1
q>0
a(n+1)^2=(an+bn)^2/(an^2+bn^2)=1+2/(an/bn+bn/an)<=2
a(n+1)<=√2
a(n+1)^2>1
a(n+1)>1
又{an}为等比数列
若0<q<1
n>=[log(q)(1/a1)]时，a(n+1)<1
不符合
若q>1
n>=[log(q)(√2/a1)]时，a(n+1)>√2
亦不符合
于是q=1
b(n+1)=√2*bn/an=√2*bn/a1
a(n+1)=[(an+bn)/(√(an^2+bn^2))]=(1+bn/an)/(√(1+(bn/an)^2))
=(1+bn/a1)/(√(1+(bn/a1)^2))=a1
则a1/bn+bn/a1为定值
又b(n+1)/bn=√2/a1为定值
{bn}必为单调数列
故而bn为定值
b(n+1)=bn=√2*bn/a1
a1=√2
a2=a1=(a1+b1)/(√(a1^2+b1^2))
b1=√2
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