Let $\mathcal G \left({z}\right) = \displaystyle \sum_{n \mathop \ge 0} \mathop F_n z^n$ be a generating function for $\mathcal F_n$. Then we have: {{begin-eqn}} {{eqn | l = \left({1 - z - z^2}\right) \mathcal G \left({z}\right)       | r = \left({\mathcal F_0 + \mathcal F_1 z + \mathcal F_2 z^2 + \mathcal F_3 z^3 + \mathcal F_4 z^4 + \cdots}\right)       | c =  }} {{eqn | o =        | ro= -       | r = \left({\mathcal F_0 z + \mathcal F_1 z^2 + \mathcal F_2 z^3 + \mathcal F_3 z^4 + \mathcal F_4 z^5 + \cdots}\right)       | c =  }} {{eqn | o =        | ro= -       | r = \left({\mathcal F_0 z^2 + \mathcal F_1 z^3 + \mathcal F_2 z^4 + \mathcal F_3 z^5 + \mathcal F_4 z^6 + \cdots}\right)       | c =  }} {{eqn | r = \mathcal F_0 + \left({\mathcal F_1 - \mathcal F_0}\right) z + \left({\mathcal F_2 - \mathcal F_1 - \mathcal F_0}\right) z^2 + \left({\mathcal F_3 - \mathcal F_2 - \mathcal F_1}\right) z^3 + \cdots       | c =  }} {{eqn | r = \mathcal F_0 + \left({\mathcal F_1 - \mathcal F_0}\right) z + F_0 z^2 + F_1 z^3 + \cdots       | c = {{Defof|Second Order Fibonacci Number}}: $\mathcal F_n - \mathcal F_{n - 1} - \mathcal F_{n - 2} = F_{n - 2}$ }} {{eqn | r = z + z^2 \sum_{k \mathop \ge 0} F_k z^k       | c = {{Defof|Second Order Fibonacci Number}}: $\mathcal F_0 = 0$, $\mathcal F_1 = 1$ }} {{eqn | r = z + z^2 G \left({z}\right)       | c = where $G \left({z}\right)$ is a generating function for the Fibonacci numbers }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \mathcal G \left({z}\right)       | r = \dfrac {z + z^2 G \left({z}\right)} {1 - z - z^2}       | c =  }} {{eqn | r = \dfrac z {1 - z - z^2} + \dfrac z {1 - z - z^2} z G \left({z}\right)       | c =  }} {{eqn | r = G \left({z}\right) + z \left({G \left({z}\right)}\right)^2       | c = Generating Function for Fibonacci Numbers: $G \left({z}\right) = \dfrac z {1 - z - z^2}$ }} {{end-eqn}} Then from Summation over k to n of Product of kth with n-kth Fibonacci Numbers, the coefficient of $z^n$ in $\left({G \left({z}\right)}\right)^2$ is: :$\dfrac {\left({n - 1}\right) F_n + 2n F_{n - 1} } 5$ Thus the coefficient of $z^{n + 1}$ in $z \left({G \left({z}\right)}\right)^2$ is likewise: :$\dfrac {\left({n - 1}\right) F_n + 2n F_{n - 1} } 5$ and so the coefficient of $z^n$ in $G \left({z}\right) + z \left({G \left({z}\right)}\right)^2$ is: :$F_n + \dfrac {\left({n - 2}\right) F_{n - 1} + 2 \left({n - 1}\right) F_{n - 2} } 5$ Hence: {{begin-eqn}} {{eqn | l = F_n + \dfrac {\left({n - 2}\right) F_{n - 1} + 2 \left({n - 1}\right) F_{n - 2} } 5       | r = F_n + \dfrac {\left({2 n - 2}\right) F_{n - 1} - n F_{n - 1} + \left({2 n - 2}\right) F_{n - 2} } 5       | c =  }} {{eqn | r = F_n + \dfrac {\left({2 n - 2}\right) F_n - n F_{n - 1} } 5       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = \dfrac {5 F_n + \left({2 n - 2}\right) F_n - n F_{n - 1} } 5       | c = common denominator }} {{eqn | r = \dfrac {\left({2 n + 3}\right) F_n - n F_{n - 1} } 5       | c =  }} {{eqn | r = \dfrac {\left({2 n + 3}\right) F_n + n F_n - n F_n - n F_{n - 1} } 5       | c =  }} {{eqn | r = \dfrac {\left({2 n + 3}\right) F_n + n F_n - n \left({F_n + F_{n - 1} }\right)} 5       | c =  }} {{eqn | r = \dfrac {\left({2 n + 3}\right) F_n + n F_n - n F_{n + 1} } 5       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = \dfrac {\left({3 n + 3}\right) F_n} 5 - \dfrac {n F_{n + 1} } 5       | c = simplifying }} {{end-eqn}} {{qed}}