My way of finding closed form Fibonacci numbers

It seems impossible to use repeated subsitution.
Then do it mathematics way.

sqrt():square root of a number.

Assume  H(1)=a1=1 H(2)=a2=1

              H(n)=a_n=a_n-1+a_n-2

   Suppose a_n+αa_n-1=β(a_n-1+αa_n-2)

         When a_n=(β-α)a_n-1+αβa_n-2

         Compare two equation and get

        β-α=1

         αβ=1

 suppose α>0, β>0

      α=(sqrt(5)-1)/2

      β=(sqrt(5)+1)/2

since a_n+αa_n-1=β(a_n-1+αa_n-2) it is geometric progression.

So a_n+1+αa_n =( a₁+αa₂)β^n-1

                 =(1+α) β^n-1

                  =βⁿ

Suppose b_n=a_n/βⁿ

And suppose b_n+1+λ=-α/β* (b_n +λ)

λ=-1/(α+β)

since b_n+λ=(-α/β)^n-1(b₁+λ)

then b_n+λ=(-α/β)^n-1(1/β+λ)

then get a_n=sqrt(5)/5 *{[(sqrt(5)-1)/2]ⁿ+[(sqrt(5)+1)/2]ⁿ}

then H(n)= a_n=sqrt(5)/5 *{[(sqrt(5)-1)/2]ⁿ+[(sqrt(5)+1)/2]ⁿ}

It was so hard to calculate so I put it on blog in order to learn it by heart.