One confusing step in a solution

**stevey** · September 15th 2010, 05:03 PM

Hi,

I'm trying to understand the solution to a problem, and I get most of it, but there is one step in the algebra I can't quite follow. Please help!

**pickslides** · September 15th 2010, 05:31 PM

It looks like the $\frac{1+\sqrt{5}}{2}$ and the $\frac{1-\sqrt{5}}{2}$ terms have been grouped together, then factored.

**Soroban** · September 15th 2010, 06:44 PM

Hello, stevey!

pockslides is correct . . .

$\displaystyle{\frac{1}{\sqrt{5}}\left(\frac{1+\sqr t{5}}{2}\right)^{n-1}\!\!\! - \frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{n-1} \!\!\!+ \frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right )^2 \!- \frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^n$

. . $\displaystyle =\;\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\ri ght)^{n-1}\!\!\left(\frac{1+\sqrt{5}}{2}+1\right) - \frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{n-1}\!\!\left(\frac{1-\sqrt{5}}{2} + 1\right)$

Watch very carefully . . .

$\displaystyle{\underbrace{\frac{1}{\sqrt{5}}\left( \frac{1+\sqrt{5}}{2}\right)^{n-1}}_{A}\!\!\! - \underbrace{\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{n-1}}_{B} \!\!\!+ \underbrace{\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{ 5}}{2}\right)^n}_{C} \!- \underbrace{\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^n}_{D}$

. . $\displaystyle =\; \underbrace{\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{ 5}}{2}\right)^n}_{C}\!\!\! + \underbrace{\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{n-1}}_{A} \!\!\! - \underbrace{\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{ 5}}{2}\right)^n}_{D} \!- \underbrace{\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^n}_{B}$

. . $\displaystyle =\; \overbrace{\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5 }}{2}\right)^{n-1}\!\!\left(\frac{1+\sqrt{5}}{2} + 1\right)}^{\text{Factor }C\text{ and }A} - \overbrace{\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{n-1}\!\!\left(\frac{1-\sqrt{5}}{2} + 1\right)}^{\text{Factor }D\text{ and }B}$

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