pure難題一問

given {a(n)} is a sequence which defined bya(1)=1 and a(2)=3 and
a(n+2)=a(n+1)+a(n) without using mathematical induction
prove that a(n)<(7/4)^n for n=1，2，3，.....
Sol
a(n+2)=a(n+1)+an
E^2=E+1
E^2－E－1=0
E=(1+/-√5)/2
S=(1+√5)/2，T=(1－√5)/2
S－T=√5
a(n)=p*S^n+q*T^n
a(n+2)=a(n+1)+a(n)
a(2)=a(1)+a(0)
a(0)=a(2)－a(1)=3－1=2
a(0)=2=p+q
a(1)=1=pS+qT
S(p+q)－(pS+qT)=2S－1
q(S－T)=2S－1
q*√5=√5
q=1，p=1
a(n)=[(1+√5)/2]^n+[(1－√5)/2]^n
～1.618^n+(－0.618)^n
<1.75^n
=(7/4)^n
So
a(n)<(7/4)^n

當n為偶數時,證明時不應這樣簡單?