i) Write down the characteristic equation:
x2=x+1
x2−x−1=0
x1=21+5=ϕ; x2=21−5=ψ
Fn=ax1n+bx2n
a∗ϕ+b∗ψ=1 and a∗ϕ2+b∗ψ2=1
a∗ϕ=1−b∗ψ;
(1−b∗ψ)∗ϕ+b∗ψ2=1
ϕ∗ψ=41−5=−1;
ϕ+b+b∗ψ2=1
b∗(1+ψ2)=1−ϕ
ψ2>0; 1+ψ2>0
b=1+ψ21−ϕ=1+(21−5)21−21+5=
=42∗5−2∗521−5=−5∗2(1−5)21−5=−51
ϕ>0; a=ϕ1+5ψ=21+51+521−5=
=21+5252∗5+1−5=21+5251+5=
=51∗21+521+5=51
Fn=5ϕn−5ψn=5ϕn−ψn;
ii) a3=2∗5−1+2=10+1=11
an+1=2an−an−1+2;
an=2an−1−an−2+2;
From the first equation subtract the second:
an+1−an=2an−an−1+2−2an−1+an−2−2
an+1=3an−3an−1+an−2 for n≥3
Therefore, for n≥4:an=3an−1−3an−2+an−3;
Write down the characteristic equation:
x3−3x2+3x−1=0
(x−1)3=0
x1,2,3=1;
an=k1∗1n+k2∗n∗1n+k3∗n2∗1n=
=k1+k2∗n+k3∗n2;
The system of three equations:
k1+k2+k3=a1=1
k1+k2∗2+k3∗4=a2=5
k1+k2∗3+k3∗9=a3=11
Subtract the first from the second:
k2+k3∗3=4 (iv)
Subtract the second from the third:
k2+k3∗5=6 (v)
Subtract (iv) from (v):
k3∗2=2
k3=1
From this and (iv):
k2+3=4
k2=1
From these and the first:
k1+1+1=1
k1=−1
an=−1+n+n2=n2+n−1.
Answer: i) Fn=5ϕn−ψn
ii) an=n2+n−1