Mathematics Q

F(n) = [(1+square root 5)/2]^n + [(1-square root 5)/2]^n
F(n-1) = [(1+square root 5)/2]^(n-1) + [(1-square root 5)/2]^(n-1)
= [(1+sqrt 5)/2]^n*[2/(1+sqrt 5)] + [(1-sqrt 5)/2]^n*[2/(1-sqrt 5)]

F(n)+F(n-1)=
[(1+sqrt 5)/2]^n*[1+2/(1+sqrt 5)] + [(1-sqrt 5)/2]^n*[1+2/(1-sqrt 5)]

since
1+2/(1+sqrt 5)
=1+2(1-sqrt5)/(1-5)
=1-(2-2sqrt5)/4
=(4-2+2sqrt5)/4
=(2+2sqrt5)/4
=(1+sqrt5)/2

and
1+2/(1-sqrt 5)
=1+2(1+sqrt5)/(1-5)
=1-(2+2sqrt5)/4
=(4-2-2sqrt5)/4
=(2-2sqrt5)/4
=(1-sqrt5)/2

so:
F(n)+F(n-1)=
[(1+sqrt 5)/2]^n*[(1+sqrt5)/2] + [(1-sqrt 5)/2]^n*[(1-sqrt5)/2]
=[(1+square root 5)/2]^(n+1) + [(1-square root 5)/2]^(n+1)

=LHS
=F(n+1)
= [(1+square root 5)/2]^(n+1) + [(1-square root 5)/2]^(n+1)

Then
Put n=0,
F(0)=2

Put n=1,
F(1)=1

F(2)=0+1=3
F(3)=1+3=4
F(4)=3+4=7
F(5)=4+7=11
F(6)=7+11=18
F(7)=11+18=29
F(8)=18+29=47