# Let displaystyle{F}_{{i}} be in the displaystyle{i}^{{{t}{h}}} Fibonacc number, and let n be ary positive eteger displaystylege{3} Prove that displaystyle{F}_{{n}}=frac{1}{{4}}{left({F}_{{{n}-{2}}}+{F}_{{n}}+{F}_{{{n}+{2}}}right)}

Question
Discrete math
Let $$\displaystyle{F}_{{i}}$$ be in the $$\displaystyle{i}^{{{t}{h}}}$$ Fibonacc number, and let n be ary positive eteger $$\displaystyle\ge{3}$$
Prove that
$$\displaystyle{F}_{{n}}=\frac{1}{{4}}{\left({F}_{{{n}-{2}}}+{F}_{{n}}+{F}_{{{n}+{2}}}\right)}$$

2021-01-20
Let us first recall a definition of nth Fibonacci number
$$\displaystyle{F}_{{n}}={F}_{{{n}-{1}}}+{F}_{{{n}-{2}}},\text{for}\ {n}\ge{2}$$
Now we have to show
$$\displaystyle{F}_{{n}}=\frac{1}{{4}}{\left({F}_{{{n}-{2}}}+{F}_{{n}}+{F}_{{{n}+{2}}}\right)},\text{for}\ {n}\ge{3}$$
Now starting from right hand side we get
$$\displaystyle\frac{1}{{4}}{\left({F}_{{{n}-{2}}}+{F}_{{n}}+{F}_{{{n}+{2}}}\right)}=\frac{1}{{4}}{\left({F}_{{{n}-{2}}}+{F}_{{n}}+{\left({F}_{{{n}+{1}}}{F}_{{n}}\right)}\right)}{\left[\therefore{F}_{{{n}+{2}}}={F}_{{{n}+{1}}}+{F}_{{n}}\right]}$$
$$\displaystyle=\frac{1}{{4}}{\left({F}_{{{n}-{2}}}+{2}{F}_{{n}}+{F}_{{{n}+{1}}}\right)}{\left[\therefore{F}_{{{n}+{2}}}={F}_{{{n}+{1}}}+{F}_{{n}}\right]}$$
$$\displaystyle=\frac{1}{{4}}{\left({F}_{{{n}-{2}}}+{2}{F}_{{n}}+{\left({F}_{{n}}{F}_{{{n}-{1}}}\right)}\right)}{\left[\therefore{F}_{{{n}+{1}}}={F}_{{n}}+{F}_{{{n}-{1}}}\right]}$$
$$\displaystyle=\frac{1}{{4}}{\left({\left({F}_{{{n}-{2}}}+{F}_{{{n}-{1}}}\right)}{3}{F}_{{n}}\right)} \(\displaystyle=\frac{1}{{4}}{\left({4}{F}_{{n}}\right)}{\left[\therefore{F}_{{n}}={F}_{{{n}-{1}}}+{F}_{{{n}-{2}}}\right]}$$
$$\displaystyle={F}_{{n}}$$
Hence the proved

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...