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I have no idea how to do this.

If $a = \dfrac{\sqrt{5}+1}{2}$ then $(a+1)^{25} =$?

I tried to transform $a$ to polar form but failed, because if it can be in polar form, I could calculate with de Moivre's formulae.

Can anyone please explain me how to do this easily?

Thank you

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  • $\begingroup$ Talking complex numbers, where is the $i$ ? $\endgroup$
    – user138999
    Jul 14, 2014 at 12:24
  • $\begingroup$ It's a positive real number, so polar form doesn't change anything. $\endgroup$
    – hardmath
    Jul 14, 2014 at 12:25
  • $\begingroup$ maybe its $\sqrt{5}+i$? $\endgroup$
    – cirpis
    Jul 14, 2014 at 12:26
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    $\begingroup$ Note that $a^2=a+1$, $a^3=2a+1$, $a^4=3a+2$, and so on... (Hint: Fibonacci numbers). $\endgroup$
    – Oleg567
    Jul 14, 2014 at 12:31
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    $\begingroup$ Approximately $28143753123$. $\endgroup$ Jul 14, 2014 at 12:58

2 Answers 2

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Note that the number you are refering to is the golden ratio $a=\phi=\frac{1+\sqrt{5}}{2}$, and it is a solution to the equation $$\phi^2-\phi-1=0$$ thus rearanging we get $$\phi^2=\phi+1$$ multiply by $\phi$ to get $$\phi^3=\phi^2+\phi=2\phi+1$$ and again $$\phi^4=2\phi^2+\phi=3\phi+2$$ and again $$\phi^5=3\phi^2+2\phi=5\phi+3$$ until you start to see the pattern of Fibonacci numbers in the coeficients, thus for any natural $n$ $$\phi^n=F_n\phi+F_{n-1}$$ where $F_n$ is the nth Fibonacci number.

Then, since $\phi^2=\phi+1$ we get that $$(\phi+1)^{25}=\phi^{50}=F_{50}\phi+F_{49}$$

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Hint:

Note that
$a^2=a+1$;
$a^3=2a+1$;
$a^4=3a+2$;
$a^5=5a+3$;
$...$
$a^n = F_{n}a+F_{n-1}$,
where $F_n$ is $n$-th Fibonacci number.

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