f(54) = (1+sqrt(5))^54-(1-sqrt(5))^54/2^54*sqrt(5)

Thanks, Alan...this is derived from "Binet's Formula" which allows the calculation of any Fibonacci number, (n).

The "formula" is given by:

F(n) = [(Phi)ⁿ - (-phi)ⁿ] / √5

Where      Phi = [√5 + 1] / 2      and     phi  =  [√5 - 1] / 2       So we have........

F(n) = [ ((√5 + 1) / 2)ⁿ -  ((1- √5 ) / 2)ⁿ ] / √5 =  [ (1 + √5 )ⁿ -  (1- √5 )ⁿ ] / (2ⁿ√5)

This webpage has some fascinating facts and properties of the Fibonacci Series.......it's one of my favorite math sites on the internet!!!

f(54) = (1+sqrt(5))^54-(1-sqrt(5))^54/2^54*sqrt(5)  -------?

$$\begin{array}{rll}
f(54)&=&(1+\sqrt5)^{54}-\frac{(1-\sqrt5)^{54}}{2^{54}}\times\sqrt5
\end{array}$$

Before I even think about going further, is this your intended question?

I suspect it should be in the following form (where n is a positive integer):

$$f(n)=\frac{(1+\sqrt5)^n-(1-\sqrt5)^n}{2^n \sqrt5}$$

This function generates the Fibonacci sequence, so the questioner is asking for the 54'th Fibonacci number.

$${\frac{\left({\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{5}}}}\right)}^{{\mathtt{54}}}{\mathtt{\,-\,}}{\left({\mathtt{1}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{5}}}}\right)}^{{\mathtt{54}}}\right)}{\left({{\mathtt{2}}}^{{\mathtt{54}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}\right)}} = {\mathtt{86\,267\,571\,271.999\: \!999\: \!999\: \!756\: \!697\: \!1}}$$

The calculator here has some problems with precision!  The answer should be 86267571272.

Just to illustrate that the function really does generate the numbers in the Fibonacci sequence here are the first 20 values calculated from it (n = 1, 2, 3, ..., 20) (from the third one onwards each one is the sum of the previous two).