Fun question...... :)

Evaluate 

\(\dfrac{1}{\sqrt5}\cdot\left(\left(\dfrac{1+\sqrt5}{2}\right)^5-\left(\dfrac{1-\sqrt5}{2}\right)^5\right)\)

And DO NOT use a calculator.

The nth element in the Fibonacci series as an analytic function of n is:

\({\displaystyle F_{n}={\cfrac {1}{\sqrt {5}}}\left({\cfrac {1+{\sqrt {5}}}{2}}\right)^{n}-{\cfrac {1}{\sqrt {5}}}\left({\cfrac {1-{\sqrt {5}}}{2}}\right)^{n}~} \)

For n=5 we calculate F₅

\(\begin{array}{|rcll|} \hline F_0 &=& 0 \\ F_1 &=& 1 \\ F_2 &=& F_0 +F_1 = 0+1= 1 \\ F_3 &=& F_1 +F_2 = 1+1= 2 \\ F_4 &=& F_2 +F_3 = 1+2= 3 \\ \mathbf{F_5} &=& F_3 +F_4 = 2+3 \mathbf{= 5} \\ \hline \end{array} \)

Mmm

\(\dfrac{1}{\sqrt5}\cdot\left(\left(\dfrac{1+\sqrt5}{2}\right)^5-\left(\dfrac{1-\sqrt5}{2}\right)^5\right)\)

\((1+\sqrt5)^5-(1-\sqrt5)^5\\ =1+5(\sqrt5)+10(\sqrt5)^2+10(\sqrt5)^3+5(\sqrt5)^4+(\sqrt5)^5\\ -(1+5(-\sqrt5)+10(-\sqrt5)^2+10(-\sqrt5)^3+5(-\sqrt5)^4+(-\sqrt5)^5)\\~\\ =+5(\sqrt5)+10(\sqrt5)^3+(\sqrt5)^5\\ -(+5(-\sqrt5)+10(-\sqrt5)^3+(-\sqrt5)^5)\\~\\ =10\sqrt5+20(\sqrt5)^3+2(\sqrt5)^5\\ =10\sqrt5+100(\sqrt5)+50(\sqrt5)\\ =160\sqrt5 \)

SO

\(\frac{1}{\sqrt5}\cdot (\frac{160\sqrt5}{32})\\ =\frac{160}{32}\\ =5\)

(you forgot your signature Max LOL)

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