A_n =(1+(√5))^n −(1−(√5))^n A


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Question Number 124041 by pticantor last updated on 30/Nov/20

$A_{n} = {(1 + \sqrt{5})}^{n} - {(1 - \sqrt{5})}^{n}$ $A_{n} = ? ? ? ?$
Commented by Dwaipayan Shikari last updated on 30/Nov/20

$A_{n} = {(1 + \sqrt{5})}^{n} - {(1 - \sqrt{5})}^{n}$ $\frac{A_{n}}{2^{n} \sqrt{5}} = \frac{1}{\sqrt{5}} ({(\frac{1 + \sqrt{5}}{2})}^{n} - {(\frac{1 - \sqrt{5}}{2})}^{n})$ $\Rightarrow \frac{A_{n}}{2^{n} \sqrt{5}} = F_{n} \Rightarrow A_{n} = 2^{n} \sqrt{5} F_{n} (F_{n} = n t h F i b b o n o c c i N u m b e r)$ $F i b b o n o c c i s e q u e n c e$ $1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2594, 4191, 6785 . . . .$ $A_{n} s e q u e n c e$ $2 \sqrt{5}, 4 \sqrt{5}, 16 \sqrt{5}, 48 \sqrt{5}, 160 \sqrt{5}, 512 \sqrt{5}, 832 \sqrt{5}, 2688 \sqrt{5}, 8704 \sqrt{5}, . . .$
	Answered by Bird last updated on 30/Nov/20
	$A_n =Σ_(k=0) ^n C_n ^k ((√5))^k −Σ_(k=0) ^n C_n ^k (−1)^k ((√5))^k =Σ_(k=0) ^n C_n ^k {1−(−1)^k }((√5))^k =2Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) ((√5))^(2p+1) =2(√5) Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) 5^p$
	$A_{n} = \sum_{k = 0}^{n} C_{n}^{k} {(\sqrt{5})}^{k} - \sum_{k = 0}^{n} C_{n}^{k} {(- 1)}^{k} {(\sqrt{5})}^{k}$ $= \sum_{k = 0}^{n} C_{n}^{k} {1 - {(- 1)}^{k}} {(\sqrt{5})}^{k}$ $= 2 \sum_{p = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 p + 1} {(\sqrt{5})}^{2 p + 1}$ $= 2 \sqrt{5} \sum_{p = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 p + 1} 5^{p}$
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