已知数列{an}满足a1=1,a2=2,a3=3,a4=4,a5=5,当n≥5时,a(n+1)=a1a2……an-1,若数列{bn}(n∈N*)当n≥5时,a(n+1)=a1a2……an-1,若数列{bn}(n∈N*),满足bn=a1a2……an-a1^2-a2^2-……-an^2 求证:仅存在两个正整数m,使得bm=0

问题描述:

已知数列{an}满足a1=1,a2=2,a3=3,a4=4,a5=5,当n≥5时,a(n+1)=a1a2……an-1,若数列{bn}(n∈N*)
当n≥5时,a(n+1)=a1a2……an-1,若数列{bn}(n∈N*),满足bn=a1a2……an-a1^2-a2^2-……-an^2
求证:仅存在两个正整数m,使得bm=0

当m=1时,b1=0.
显然an>0,m≠n时,am≠an.
当m=1时,b1=a1-a1^2=1-1=0.
当m>1时,
bm=a1a2……am-(a1^2+a2^2+……+am^2)
=-a1a2……am所以仅存在m=1时,使bm=0.
供参考.

b1=a1-a1^2=0,b2=-3,b3=-8,b4=-6,b5=65n≥5时,a(n+1)=a1a2……an-1 ,a6=a1a2a3a4a5-1,a6-a1a2a3a4a5=1b6=(a1a2...a5a6-a1^2-a2^2-a3^2-...-a^5^2-a6^2=(a1a2a3a4a5)^2-2a1a2a3a4a5+(a1a2a3a4a5-a1^2-a2^2-a3^2-a4^2-...