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

I want to share an Induction method to do the same:

First Step of induction is left.

Inductive Step:

\Sigma C(n+1,k) = \Sigma (C(n , k)) + \Sigma C(n , k-1) ;

since , C(n+1,k) = C(n , k) + C(n , k-1) ;

Which is equal to fm + fm-1;
so fm+1 = fm + fm-1 .

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Please confirm it from teachers before using it, I am not completely sure about, just came in mind. Seemed Good , So Shared.

@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.