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

\(\displaystyle{F}_{{n}}={F}_{{{n}-{1}}}+{F}_{{{n}-{2}}},\text{for}\ {n}\ge{2}\)

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