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

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

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Let us first recall a definition of nth Fibonacci number

Now we have to show

Now starting from right hand side we get
$\frac{1}{4}\left({F}_{n-2}+{F}_{n}+{F}_{n+2}\right)=\frac{1}{4}\left({F}_{n-2}+{F}_{n}+\left({F}_{n+1}{F}_{n}\right)\right)\left[\therefore {F}_{n+2}={F}_{n+1}+{F}_{n}\right]$
$=\frac{1}{4}\left({F}_{n-2}+2{F}_{n}+{F}_{n+1}\right)\left[\therefore {F}_{n+2}={F}_{n+1}+{F}_{n}\right]$
$=\frac{1}{4}\left({F}_{n-2}+2{F}_{n}+\left({F}_{n}{F}_{n-1}\right)\right)\left[\therefore {F}_{n+1}={F}_{n}+{F}_{n-1}\right]$
$=\frac{1}{4}\left(\left({F}_{n-2}+{F}_{n-1}\right)3{F}_{n}\right)$
$=\frac{1}{4}\left(4{F}_{n}\right)\left[\therefore {F}_{n}={F}_{n-1}+{F}_{n-2}\right]$
$={F}_{n}$
Hence the proved