Question

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

Answer 1

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
\(\displaystyle\frac{1}{{4}}{\left({F}_{{{n}-{2}}}+{F}_{{n}}+{F}_{{{n}+{2}}}\right)}=\frac{1}{{4}}{\left({F}_{{{n}-{2}}}+{F}_{{n}}+{\left({F}_{{{n}+{1}}}{F}_{{n}}\right)}\right)}{\left[\therefore{F}_{{{n}+{2}}}={F}_{{{n}+{1}}}+{F}_{{n}}\right]}\)
\(\displaystyle=\frac{1}{{4}}{\left({F}_{{{n}-{2}}}+{2}{F}_{{n}}+{F}_{{{n}+{1}}}\right)}{\left[\therefore{F}_{{{n}+{2}}}={F}_{{{n}+{1}}}+{F}_{{n}}\right]}\)
\(\displaystyle=\frac{1}{{4}}{\left({F}_{{{n}-{2}}}+{2}{F}_{{n}}+{\left({F}_{{n}}{F}_{{{n}-{1}}}\right)}\right)}{\left[\therefore{F}_{{{n}+{1}}}={F}_{{n}}+{F}_{{{n}-{1}}}\right]}\)
\(=\frac{1}{{4}}{\left({\left({F}_{{{n}-{2}}}+{F}_{{{n}-{1}}}\right)}{3}{F}_{{n}}\right)}\)
\(\displaystyle=\frac{1}{{4}}{\left({4}{F}_{{n}}\right)}{\left[\therefore{F}_{{n}}={F}_{{{n}-{1}}}+{F}_{{{n}-{2}}}\right]}\)
\(\displaystyle={F}_{{n}}\)
Hence the proved