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General Fibonacci sequence

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Post Thu Dec 08, 2016 12:56 pm

Posts: 99
The general Fibonacci sequence is defined by
\[{G_{n + 2}} = a{G_{n + 1}} + {G_n},\;n \ge 0,\;\;{G_0} = 0,\;{G_1} = 1.\]
Prove the following results:
\[\begin{array}{l}
G_n^2 - {G_{n - 1}}{G_{n + 1}} = {\left( { - 1} \right)^{n - 1}}, \\
{G_m}{G_n} + {G_{m + 1}}{G_{n + 1}} = {G_{m + n + 1}}, \\
G_n^2 + G_{n + 1}^2 = {G_{2n + 1}}, \\
G_{n + 1}^2 - G_{n - 1}^2 = a{G_{2n}}. \\
\end{array}\]

Post Wed Apr 05, 2017 9:01 am

Posts: 4
Location: South Korea
Note that we have
$$\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n = \begin{pmatrix} G_{n+1} & G_{n} \\ G_{n} & G_{n-1} \end{pmatrix}\tag{1}$$
Taking the determinant of each side in $(1)$, we have $G_{n-1}G_{n+1}-G_{n}^2=(-1)^{n}$ as desired.

The second identity can be derived from the fact that $A^{n}A^{m}=A^{n+m}$ for any square matrix $n$.

The third identity follows from putting $m=n$ in the second identity.

The fourth follows because we have (from the third identity) $$G_{n}^2+G_{n+1}^2=G_{2n+1} \tag{2}$$ $$G_{n}^2+G_{n-1}^2=G_{2n-1} \tag{3}$$

$(2)-(3)$ gives the fourth identity as $G_{2n+1}=aG_{2n}+G_{2n-1}$ and $G_{n+1}^2-G_{n-1}^2=G_{2n+1}-G_{2n-1}$

Done!
$$\int_{0}^{1}\sin{(\pi x)}x^x(1-x)^{1-x} \; \mathrm{d}x=\dfrac{\pi e}{4!}$$


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