The Golden ratio! 2

Algebra Level 3

If $\phi = \dfrac{1+\sqrt5}2$ , compute $\dfrac1\phi + \dfrac1{\phi^3}+ \dfrac1{\phi^5}+ \dfrac1{\phi^7} + \cdots$ .

\frac \phi2

\phi

\phi ^2

1 solution

Mateus Gomes
Feb 6, 2016

For a geometric progression with initial term $a$ and common ratio $r$ satisfying $|r|<1,$ the sum of the infinite terms of the geometric progression is $S_{\infty}=\frac{a}{1-r}$ $\rightarrow a=\frac{1}{\phi}, r=\frac{1}{{\phi}^2}$ $\rightarrow\frac{\frac{1}{\phi}}{1-\frac{1}{\phi^2}}=\frac{\frac{1}{\phi}}{\frac{{\phi}^{2}-1}{{\phi}^{2}}}=\frac{\phi}{{\phi}^{2}-1}=\frac{\phi}{(\phi-1)(\phi+1)}=\frac{\frac{1+\sqrt5}{2}}{(\frac{1+\sqrt5}{2}-1)(\frac{1+\sqrt5}{2}+1)}=\frac{\frac{1+\sqrt5}{2}}{(\frac{\sqrt5-1}{2})(\frac{\sqrt5+3}{2})}=(\frac{1+\sqrt5}{2})\times(\frac{2}{\sqrt5-1})\times(\frac{2}{\sqrt5+3})=(\frac{2+2\sqrt5}{2+2\sqrt5})=\Large\color{#3D99F6}{\boxed{1}}$

Nice(+1)..Did the same way.... But instead of putting values of $\phi$ you could have used the fact that $\phi^2-\phi-1=0$ or $\phi=\phi^2-1$ or $\dfrac{\phi}{\phi^2-1}=1$

Rishabh Jain - 5 years, 4 months ago

The Golden ratio! 2

1 solution

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