Answer
a) See proof
b) See proof
c) See proof
Work Step by Step
We are given:
$u_n=\dfrac{(1+\sqrt 5)^n-(1-\sqrt 5)^n}{2^n\sqrt 5}$
a) Compute $u_1$:
$u_1=\dfrac{(1+\sqrt 5)^1-(1-\sqrt 5)^1}{2^1\sqrt 5}=\dfrac{1+\sqrt 5-1+\sqrt 5}{2\sqrt 5}$
$=\dfrac{2\sqrt 5}{2\sqrt 5}=1$
Compute $u_2$:
$u_2=\dfrac{(1+\sqrt 5)^2-(1-\sqrt 5)^2}{2^2\sqrt 5}=\dfrac{1+2\sqrt 5+5-1+2\sqrt 5-5}{4\sqrt 5}$
$=\dfrac{4\sqrt 5}{4\sqrt 5}=1$
b) Compute $u_{n+1}+u_n$:
$u_{n+1}+u_n=\dfrac{(1+\sqrt 5)^{n+1}-(1-\sqrt 5)^{n+1}}{2^{n+1}\sqrt 5}+\dfrac{(1+\sqrt 5)^n-(1-\sqrt 5)^n}{2^n\sqrt 5}$
$=\dfrac{(1+\sqrt 5)^n (1+\sqrt 5)-(1-\sqrt 5)^n(1-\sqrt 5)}{2\cdot 2^n\sqrt 5}+\dfrac{(1+\sqrt 5)^n-(1-\sqrt 5)^n}{2^n\sqrt 5}$
$=\dfrac{(1+\sqrt 5)^n (1+\sqrt 5)-(1-\sqrt 5)^n(1-\sqrt 5)+2(1+\sqrt 5)^n+2(1-\sqrt 5)^n}{2\cdot 2^n\sqrt 5}$
$=\dfrac{(1+\sqrt 5)^n}{2^n\sqrt 5}\left(\dfrac{1+\sqrt 5}{2}+1 \right)-\dfrac{(1-\sqrt 5)^n}{2^n\sqrt 5}\left(\dfrac{1-\sqrt 5}{2}+1 \right)$
$=\dfrac{(1+\sqrt 5)^n}{2^n\sqrt 5}\left(\dfrac{3+\sqrt 5}{2} \right)-\dfrac{(1-\sqrt 5)^n}{2^n\sqrt 5}\left(\dfrac{3-\sqrt 5}{2} \right)$
$u_{n+2}=\dfrac{(1+\sqrt 5)^{n+2}-(1-\sqrt 5)^{n+2}}{2^{n+2}\sqrt 5}$
$=\dfrac{(1+\sqrt 5)^n(1+\sqrt 5)^2-(1-\sqrt 5)^n(1-\sqrt 5)^2}{2^{n+2}\sqrt 5}$
$=\dfrac{(1+\sqrt 5)^n(6+2\sqrt 5)-(1-\sqrt 5)^n(6-2\sqrt 5)}{2^{n+2}\sqrt 5}$
$=\dfrac{(1+\sqrt 5)^n}{2^n\sqrt 5}\left(\dfrac{6+2\sqrt 5}{4} \right)-\dfrac{(1-\sqrt 5)^n}{2^n\sqrt 5}\left(\dfrac{6-2\sqrt 5}{4} \right)$
$=\dfrac{(1+\sqrt 5)^n}{2^n\sqrt 5}\left(\dfrac{3+\sqrt 5}{2} \right)-\dfrac{(1-\sqrt 5)^n}{2^n\sqrt 5}\left(\dfrac{3-\sqrt 5}{2} \right)$
We got that $u_{n+2}=u_{n+1}+u_n$
c) As each term of the sequence is the sum of the previous two terms, the sequence $\{u_n\}$ is a Fibonacci sequence.
