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Question Number 157343 by mathocean1 last updated on 22/Oct/21

	Answered by mindispower last updated on 23/Oct/21
	$X^2 −X−1=0 ⇒X∈{((1+(√5))/2),((1−(√5))/2)} u_n =a(((1+(√5))/2))^n +b(((1−(√5))/2))^n (a,b) { ((a+b=0)),(((1/2)(a+b)+((√5)/2)(a−b)=1)) :} ⇒a=(1/( (√5))),b=−(1/( (√5))) ∅_n =(1/( (√5)))((((1+(√5))/2))^n −(((1−(√5))/2))^n ) (2)∅_(n+1) ^2 =(1/5)((((1+(√5))/2))^(2n+2) +(((1−(√5))/2))^(2n+2) −2(−1)^(n+1) ) ∅_n .∅_(n+2) =(1/5)((((1+(√5))/2))^(2n+2) +(((1−(√5))/2))^(2n+2) (−1)^n (−(((1+(√5))/2))^2 −(((1−(√5))/2))^2 ) =(1/5)((((1+(√5))/2))^(2+2n) +(((1−(√5))/2))^(2n+2) +(−1)^n (−3)) ∅_(n+1) ^2 −∅_n ∅_(n+1) =(−1)^n (3Σ)1−((∅_n /∅_(n+1) ))=(((−1)^n )/((∅_(n+1) )^2 )) l=lim_(n→∞) ((∅_(n+1) /∅_n )) 1−(1/l)=0⇒l=1 (4),Σ_(k=0) ^n C_n ^k ∅_k =(1/( (√5)))(Σ_(k=0) ^n C_n ^k (((1+(√5))/2))^k −5)Σ_(k≤n) C_n ^k (((1−(√5))/2))^(k)) =(1/( (√5)))((1+((1+(√5))/2))^n −(1+((1−(√5))/2))^n ) =(1/( (√5)))((((6+2(√5))/4))^n −(((6−2(√5))/4))^n )=(1/( (√5)))((((1+(√5))/2))^(2n) −(((1−(√5))/2))^(2n) )=∅_(2n) (((1+_− (√5))/2))^2 =((6+_− 2(√5))/4) (5) meme Chose Que la d1$
	$X^{2} - X - 1 = 0$ $\Rightarrow X \in {\frac{1 + \sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2}}$ $u_{n} = a {(\frac{1 + \sqrt{5}}{2})}^{n} + b {(\frac{1 - \sqrt{5}}{2})}^{n}$ $(a, b) {\begin{cases} a + b = 0 \\ \frac{1}{2} (a + b) + \frac{\sqrt{5}}{2} (a - b) = 1 \end{cases}$ $\Rightarrow a = \frac{1}{\sqrt{5}}, b = - \frac{1}{\sqrt{5}}$ $\emptyset_{n} = \frac{1}{\sqrt{5}} ({(\frac{1 + \sqrt{5}}{2})}^{n} - {(\frac{1 - \sqrt{5}}{2})}^{n})$ $(2) \emptyset_{n + 1}^{2} = \frac{1}{5} ({(\frac{1 + \sqrt{5}}{2})}^{2 n + 2} + {(\frac{1 - \sqrt{5}}{2})}^{2 n + 2} - 2 {(- 1)}^{n + 1})$ $\emptyset_{n} . \emptyset_{n + 2} = \frac{1}{5} ({(\frac{1 + \sqrt{5}}{2})}^{2 n + 2} + {(\frac{1 - \sqrt{5}}{2})}^{2 n + 2} {(- 1)}^{n} (- {(\frac{1 + \sqrt{5}}{2})}^{2} - {(\frac{1 - \sqrt{5}}{2})}^{2})$ $= \frac{1}{5} ({(\frac{1 + \sqrt{5}}{2})}^{2 + 2 n} + {(\frac{1 - \sqrt{5}}{2})}^{2 n + 2} + {(- 1)}^{n} (- 3))$ $\emptyset_{n + 1}^{2} - \emptyset_{n} \emptyset_{n + 1} = {(- 1)}^{n}$ $(3 Σ) 1 - (\frac{\emptyset_{n}}{\emptyset_{n + 1}}) = \frac{{(- 1)}^{n}}{{(\emptyset_{n + 1})}^{2}}$ $l = \lim_{n \to \infty} (\frac{\emptyset_{n + 1}}{\emptyset_{n}})$ $1 - \frac{1}{l} = 0 \Rightarrow l = 1$ $(4), \underset{k = 0}{\sum^{n}} C_{n}^{k} \emptyset_{k} = \frac{1}{\sqrt{5}} (\underset{k = 0}{\sum^{n}} C_{n}^{k} {(\frac{1 + \sqrt{5}}{2})}^{k} - 5) \sum_{k ⩽ n} C_{n}^{k} {(\frac{1 - \sqrt{5}}{2})}^{k)}$ $= \frac{1}{\sqrt{5}} ({(1 + \frac{1 + \sqrt{5}}{2})}^{n} - {(1 + \frac{1 - \sqrt{5}}{2})}^{n})$ $= \frac{1}{\sqrt{5}} ({(\frac{6 + 2 \sqrt{5}}{4})}^{n} - {(\frac{6 - 2 \sqrt{5}}{4})}^{n}) = \frac{1}{\sqrt{5}} ({(\frac{1 + \sqrt{5}}{2})}^{2 n} - {(\frac{1 - \sqrt{5}}{2})}^{2 n}) = \emptyset_{2 n}$ ${(\frac{1 \underline{+} \sqrt{5}}{2})}^{2} = \frac{6 \underline{+} 2 \sqrt{5}}{4}$ $(5) m e m e C h o s e Q u e l a d 1$
	Commented by mathocean1 last updated on 24/Oct/21

	$T h a n k s$
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