Sequences

Let us define a sequence of numbers {f_m} by f₀ = 0, f₁ = 1 and
f_m+2 = f_m+1 + f_m , m≥0.
Its quite obvious that f_m is the the m-th Fibonacci number for m≥1.
Consider the power series
A(x) = f₀ + f₁ x + f₂ x² + . . . + f_m x^m + . . .
Its quite obvious that
A(x) - x A(x) - x² A(x) = f₀ + (f₁ - f₀) x = x
Hence, we get
A(x) = x1-x-x²
This, therefore, is the generating function for the sequence {f_m}.
Hence,
f_m+1 = coefficient of x^m+1 in the expansion of A(x)
= coefficient of x^m in the expansion of 11-x-x²
On the other hand, we have
\dfrac{1}{1-x-x^2} = \dfrac{1}{1-x(1+x)} = \sum_{n\ge 0}(x(1+x))^n
(that's just the GP formula)
That's further
\sum_{n\ge 0} x^n (1+x)^n =\sum_{n\ge 0}x^n \sum_{k\ge 0}~^nC_k x^k = \sum_{n\ge 0}\sum_{k\ge 0}~^nC_k x^{n+k}
In these notations
^nC_k = 0 if k > n.
Hence,
f_{m+1} = \sum_{\substack{n\ge k\ge0\\n+k=m}}~^nC_k

That's the solution i also had(by induction), but after seeing kaymant sir's method i was trying to use Exponential Generating Function and i got

E"(x) = E'(x) + E(x) , where E is the exp gf of fibonacci numbers

From the above relation is it possible to find E(x) ?

i dont know differential equations

@Shubhodip,
Its definitely possible to solve the above differential equation. Its basically a second order differential equation with constant coefficients. The usual way is to find the eigenvalues of the given equation. The eigenvalues are the roots of characteristic equation which is obtained by replacing the n-th order derivative with λⁿ. To be more specific, in this case, the characteristic equation is
λ² - λ -1 =0
with roots λ_1,2 = 1±√52
(no surprise at that[1])
The general solution then is
E(x) = A e^λ₁x + B e^λ₂x
with A and B are constants.

Which leads to Binets Formula ;)

Thank you