Un=(1+(√2))^n show that we have p_n ∈N / U_n =(√p_n )+(√(p


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Question Number 102474 by pticantor last updated on 09/Jul/20

$U n = {(1 + \sqrt{2})}^{n}$ $s h o w t h a t w e h a v e p_{n} \in N /$ $U_{n} = \sqrt{p_{n}} + \sqrt{p_{n} + 1}$
	Answered by ~blr237~ last updated on 09/Jul/20

	$o b s e r v e t h a t \frac{1}{U_{n}} = {(- 1)}^{n} {(1 - \sqrt{2})}^{n} a n d t h a t t h e r e e x i s t$ $a_{n}, b_{n} \in N s u c h a s U_{n} = a_{n} + b_{n} \sqrt{2} a n d \frac{1}{U_{n}} = {(- 1)}^{n} (a_{n} - b_{n} \sqrt{2})$ $a n d U_{n}^{2} = U_{2 n} .$ $S o {(U_{n}^{} + \frac{1}{U_{n}})}^{2} = 2 + U_{2 n} + \frac{1}{U_{2 n}} = 2 + 2 a_{2 n}$ ${(U_{n} - \frac{1}{U_{n}})}^{2} = - 2 + U_{2 n} + \frac{1}{U_{2 n}} = - 2 + 2 a_{2 n}$ $w e h a v e a_{n + 1} + b_{n + 1} \sqrt{2} = (1 + \sqrt{2}) (a_{n} + b_{n} \sqrt{2}) l e a d t o$ $a_{n + 1} = a_{n} + 2 b_{n} . S o s u c h a s a_{1} = 1 (o d d), a_{n} i s a l w a y s o d d$ $S o t h e r e e x i s t c_{n} \in N / a_{n} = 2 c_{n} + 1$ $T h e n {(U_{n} + \frac{1}{U_{n}})}^{2} = 2 + 2 (2 c_{2 n} + 1) = 4 (c_{2 n} + 1)$ ${(U_{n} - \frac{1}{U_{n}})}^{2} = - 2 + 2 (2 c_{2 n} + 1) = 4 c_{2 n}$ $B y a d d i n g t h e t w o s q u a r e r o o t$ $U_{n} = \sqrt{c_{2 n}} + \sqrt{c_{2 n} + 1} t a k e p_{n} = c_{2 n}$
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