A calculus proof for the general term of the Fibonacci sequence Let $a_0=1$,$a_1=1$ and $a_n=a_{n-1} + a_{n-2}$ for $n \geq 2$, I would like to prove:
$$a_n=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{n + 1}- \left(\frac{1-\sqrt{5}}{2}\right)^{n + 1} \right)$$
for $n\in \mathbb{N^*}$.
I used induction to prove it (not sure if it is a correct way), is there a way to find it doing some calculus ? I couldn't come to that result.

Well there is no specific proof which makes use of calculus (meaning differentiation / integration) however there is a proof which does involve analysis in an indirect way. Here it goes.
The basic fact from analysis will be some concepts related to power series. A power series of type $\sum a_{n}x^{n}$ has a radius of convergence $R \geq 0$ such that the series is convergent if $|x| < R$ and divergent if $|x| > R$. The power series defines a continuous and infinitely differentiable function $f(x)$ for $|x| < R$. Also if two power series $\sum a_{n}x^{n}$ and $\sum b_{n}x^{n}$ are such that they are equal for all values of $x$ in some interval $(-R, R)$, then $a_{n} = b_{n}$ for all non-negative integers $n$. This last fact will be the main idea used in this proof.
Let $a_{n}$ denote $n^{\text{th}}$ Fibonacci number and consider the power series $F(x) = \sum_{n = 0}^{\infty}a_{n}x^{n}$. Clearly we can see that if this series converges in certain interval $(-R, R)$ then we have the following relations for $|x| < R$: $$F(x) = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + \cdots\tag{1}$$$$xF(x) = a_{0}x + a_{1}x^{2} + a_{2}x^{3} + \cdots \tag{2}$$$$x^{2}F(x) = a_{0}x^{2} + a_{1}x^{3} + a_{2}x^{4} + \cdots\tag{3}$$ Now these equations clearly give us $$F(x) - xF(x) - x^{2}F(x) = a_{0} + (a_{1} - a_{0})x$$ as rest of the terms cancel because of the recurrence relation $a_{n + 2} = a_{n + 1} + a_{n}$. Putting $a_{0} = 1, a_{1} = 1$ we get $$F(x) = \frac{1}{1 - x - x^{2}}$$ If $\alpha = (1 + \sqrt{5})/2, \beta = (1 - \sqrt{5})/2$ then $\alpha, \beta$ are roots of $t^{2} - t - 1 = 0$ and hence we can write $$1 - x - x^{2} = (1 - \alpha x)(1 - \beta x)$$ Thus we have $$F(x) = \frac{1}{(1 - \alpha x)(1 - \beta x)} = \frac{A}{1 - \alpha x} + \frac{B}{1 - \beta x}$$ where $A, B$ need to be found out. Clearly we have $$1 = A(1 - \beta x) + B(1 - \alpha x)$$ and putting $x = 1/\alpha$ we get $A = \dfrac{\alpha}{\alpha - \beta}$. Similarly $B = -\dfrac{\beta}{\alpha - \beta}$. Thus we have $$F(x) = \frac{1}{\alpha - \beta}\left(\frac{\alpha}{1 - \alpha x} - \frac{\beta}{1 - \beta x}\right)\tag{4}$$ Next we use another result from power series namely that $$\frac{1}{1 - x} = 1 + x + x^{2} + x^{3} + \cdots$$ for $|x| < 1$. Using this result we can see that $$\sum_{n = 0}^{\infty}a_{n}x^{n} = F(x) = \frac{1}{\alpha - \beta}\left(\sum_{n = 0}^{\infty}(\alpha^{n + 1} - \beta^{n + 1})x^{n}\right)$$ and hence the coefficients in both power series must be equal and we have $$a_{n} = \frac{\alpha^{n + 1} - \beta^{n + 1}}{\alpha - \beta} = \frac{1}{\sqrt{5}}\left\{\left(\frac{1 + \sqrt{5}}{2}\right)^{n + 1} - \left(\frac{1 - \sqrt{5}}{2}\right)^{n + 1}\right\}$$
