由于 bn=(2n-1)*[(4/5)^n]
則:b(n+1)=[2(n+1)-1]*[(4/5)^(n+1)]
=(2n+1)*[(4/5)^(n+1)]
=[(8n+4)/5]*[(4/5)^n]
則:b(n+1)-bn
=[(8n+4)/5]*[(4/5)^n]-(2n-1)*[(4/5)^n]
=[(9-2n)/5]*[(4/5)^n]
則:當(dāng)9-2n>0
,nbn
當(dāng)9-2n=9/2時(shí)
b(n+1)=5時(shí),b(n+1)