91Ó°ÊÓ

The n^thterm of the sequence is:

$u_n=\frac{{(1+\sqrt{5})}^n-{(1-\sqrt{5})}^n}{2^n\sqrt{5}}$

First term is:

$$\begin{gathered} u_1=\frac{{(1+\sqrt{5})}^1-{(1-\sqrt{5})}^1}{2^1\sqrt{5}} \\ ⇒u_1=\frac{1+\sqrt{5}-1+\sqrt{5}}{2\sqrt{5}} \\ ⇒u_1=\frac{2\sqrt{5}}{2\sqrt{5}} \\ ⇒u_1=1 \end{gathered}$$

Second term is:

$$\begin{gathered} u_1=\frac{{(1+\sqrt{5})}^2-{(1-\sqrt{5})}^2}{2^2\sqrt{5}} \\ ⇒u_2=\frac{(1+2\sqrt{5}+5)-(1-2\sqrt{5}+5)}{4\sqrt{5}} \\ ⇒u_2=\frac{4\sqrt{5}}{4\sqrt{5}} \\ ⇒u_2=1 \end{gathered}$$

Firstly, right hand side:

$$\begin{gathered} u_{n+2}=\frac{{(1+\sqrt{5})}^{n+2}-{(1-\sqrt{5})}^{n+2}}{2^{n+2}\sqrt{5}} \\ ⇒u_{n+2}=\frac{{(1+\sqrt{5})}^n{(1+\sqrt{5})}^2-{(1-\sqrt{5})}^n{(1-\sqrt{5})}^2}{2^2×2^n\sqrt{5}} \\ ⇒u_{n+2}=\frac{{(1+\sqrt{5})}^n\left[1+5+2\sqrt{5}\right]-{(1-\sqrt{5})}^n\left[1-2\sqrt{5}+5\right]}{4×2^n\sqrt{5}} \\ ⇒u_{n+2}=\frac{2×\{{(1+\sqrt{5})}^n\left[3+\sqrt{5}\right]-{(1-\sqrt{5})}^n\left[3-\sqrt{5}\right]\}}{4×2^n\sqrt{5}} \\ ⇒u_{n+2}=\frac{\{{(1+\sqrt{5})}^n\left[3+\sqrt{5}\right]-{(1-\sqrt{5})}^n\left[3-\sqrt{5}\right]\}}{2×2^n\sqrt{5}} \end{gathered}$$

Now, the left hand side:

$$\begin{gathered} u_{n+1}+u_n=\left[\frac{{(1+\sqrt{5})}^{n+1}-{(1-\sqrt{5})}^{n+1}}{2^{n+1}\sqrt{5}}\right]+\left[\frac{{(1+\sqrt{5})}^n-{(1-\sqrt{5})}^n}{2^n\sqrt{5}}\right] \\ =\frac{(1+\sqrt{5}){(1+\sqrt{5})}^n-{(1-\sqrt{5})(1-\sqrt{5})}^n+2{(1+\sqrt{5})}^n-2{(1-\sqrt{5})}^n}{2×2^n\sqrt{5}} \\ =\frac{(1+\sqrt{5}){(1+\sqrt{5})}^n+2{(1+\sqrt{5})}^n-{(1-\sqrt{5})(1-\sqrt{5})}^n-2{(1-\sqrt{5})}^n}{2×2^n\sqrt{5}} \\ =\frac{[1+\sqrt{5}+2]{(1+\sqrt{5})}^n-{[1-\sqrt{5}+2](1-\sqrt{5})}^n}{2×2^n\sqrt{5}} \\ =\frac{[3+\sqrt{5}]{(1+\sqrt{5})}^n-{[3-\sqrt{5}](1-\sqrt{5})}^n}{2×2^n\sqrt{5}} \end{gathered}$$

Since both the sides have equal results, the expression stands valid.

As proved in Step 3, the expression is a form of Fibonacci sequence. Thus, {u_n} is a Fibonacci sequence.