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Question Number 157695 by amin96 last updated on 26/Oct/21

	Answered by Rasheed.Sindhi last updated on 26/Oct/21

	$x^{16} + \frac{1}{x^{16}} = {(x^{8} + \frac{1}{x^{8}})}^{2} - 2 = 2207$ $x^{8} + \frac{1}{x^{8}} = \sqrt{2209} = 47$ $x^{8} + \frac{1}{x^{8}} = {(x^{4} + \frac{1}{x^{4}})}^{2} - 2 = 47$ $x^{4} + \frac{1}{x^{4}} = \sqrt{49} = 7$ $x^{4} + \frac{1}{x^{4}} = {(x^{2} + \frac{1}{x^{2}})}^{2} - 2 = 7$ $x^{2} + \frac{1}{x^{2}} = \sqrt{9} = 3$ $x^{2} + \frac{1}{x^{2}} = {(x + \frac{1}{x})}^{2} - 2 = 3$ $x + \frac{1}{x} = \sqrt{5}$ $. . . . . . .$ $. . . . .$
	Commented by amin96 last updated on 26/Oct/21

	I know this sir . how will it be after ?
	Commented by Rasheed.Sindhi last updated on 26/Oct/21
	Very hard to continue further sir.I have done only easy part. :)
	Answered by mr W last updated on 26/Oct/21

	${(x^{8} + \frac{1}{x^{8}})}^{2} = 2209 = 47^{2}$ $x^{8} + \frac{1}{x^{8}} = 47$ ${(x^{4} + \frac{1}{x^{4}})}^{2} = 49$ $x^{4} + \frac{1}{x^{4}} = 7$ ${(x^{2} + \frac{1}{x^{2}})}^{2} = 9$ $x^{2} + \frac{1}{x^{2}} = 3$ $l e t t = x^{2}$ $t + \frac{1}{t} = 3$ $t^{2} - 3 t + 1 = 0$ $p_{n} = t^{n} + \frac{1}{t^{n}} = x^{2 n} + \frac{1}{x^{2 n}}$ $p_{n + 1} = 3 p_{n} - p_{n - 1}$ $r^{2} - 3 r + 1 = 0$ $r_{1, 2} = \frac{3 \pm \sqrt{5}}{2}$ $p_{n} = r_{1}^{n} + r_{2}^{n} = {(\frac{3 + \sqrt{5}}{2})}^{n} + {(\frac{3 - \sqrt{5}}{2})}^{n}$ $a_{n} = {(x^{n} + \frac{1}{x^{n}})}^{2} = x^{2 n} + \frac{1}{x^{2 n}} + 2 = p_{n} + 2$ $\underset{n = 1}{\sum^{m = 2021}} a_{n} = \underset{n = 1}{\sum^{m}} (r_{1}^{n} + r_{2}^{n} + 2)$ $= \frac{r_{1} (r_{1}^{m} - 1)}{r_{1} - 1} + \frac{r_{2} (r_{2}^{m} - 1)}{r_{2} - 1} + 2 m$ $= \frac{(\frac{3 + \sqrt{5}}{2}) [{(\frac{3 + \sqrt{5}}{2})}^{m} - 1]}{\frac{3 + \sqrt{5}}{2} - 1} + \frac{(\frac{3 - \sqrt{5}}{2}) [{(\frac{3 - \sqrt{5}}{2})}^{m} - 1]}{\frac{3 - \sqrt{5}}{2} - 1} + 2 m$ $= \frac{(3 + \sqrt{5}) [{(\frac{3 + \sqrt{5}}{2})}^{m} - 1]}{\sqrt{5} + 1} - \frac{(3 - \sqrt{5}) [{(\frac{3 - \sqrt{5}}{2})}^{m} - 1]}{\sqrt{5} - 1} + 2 m$ $= (\frac{\sqrt{5} + 1}{2}) {(\frac{3 + \sqrt{5}}{2})}^{m} - (\frac{\sqrt{5} - 1}{2}) {(\frac{3 - \sqrt{5}}{2})}^{m} + 2 m - 1$
	Answered by mindispower last updated on 26/Oct/21
	$U_n =x^n +(1/x^n ),u_0 =2,u_1 =(√5) U_(n+1) =(√5)U_n −U_(n−1) X^2 −(√5)X+1=0 ⇒X∈{(((√5)−1)/2),((1+(√5))/2)} U_n =a(((1+(√5))/2))^n +b((((√5)−1)/2))^n a+b=2,(1/2)(a−b)+((√5)/2)(a+b)=(√5) a=b⇒a=b=1 U_n =(((1+(√5))/2))^n +((((√5)−1)/2))^n we want Σ_(n=1) ^(2207) (U_(2n) +2) 2.2207+Σ_(n=1) ^(2207) U_(2n) =(((1+(√5))/2))^2 .((1−(((1+(√5))/2))^(4414) )/(1−(((1+(√5))/2))^2 ))+((((√5)−1)/2))^2 .((1−((((√5)−1)/2))^(4414) )/(1−((((√5)−1)/2))^2 ))+4414$
	$U_{n} = x^{n} + \frac{1}{x^{n}}, u_{0} = 2, u_{1} = \sqrt{5}$ $U_{n + 1} = \sqrt{5} U_{n} - U_{n - 1}$ $X^{2} - \sqrt{5} X + 1 = 0$ $\Rightarrow X \in {\frac{\sqrt{5} - 1}{2}, \frac{1 + \sqrt{5}}{2}}$ $U_{n} = a {(\frac{1 + \sqrt{5}}{2})}^{n} + b {(\frac{\sqrt{5} - 1}{2})}^{n}$ $a + b = 2, \frac{1}{2} (a - b) + \frac{\sqrt{5}}{2} (a + b) = \sqrt{5}$ $a = b \Rightarrow a = b = 1$ $U_{n} = {(\frac{1 + \sqrt{5}}{2})}^{n} + {(\frac{\sqrt{5} - 1}{2})}^{n}$ $w e w a n t$ $\underset{n = 1}{\sum^{2207}} (U_{2 n} + 2)$ $2.2207 + \underset{n = 1}{\sum^{2207}} U_{2 n} = {(\frac{1 + \sqrt{5}}{2})}^{2} . \frac{1 - {(\frac{1 + \sqrt{5}}{2})}^{4414}}{1 - {(\frac{1 + \sqrt{5}}{2})}^{2}} + {(\frac{\sqrt{5} - 1}{2})}^{2} . \frac{1 - {(\frac{\sqrt{5} - 1}{2})}^{4414}}{1 - {(\frac{\sqrt{5} - 1}{2})}^{2}} + 4414$
	Answered by Raxreedoroid last updated on 28/Oct/21

	$x^{16} + \frac{1}{x^{16}} = {(x^{8} + \frac{1}{x^{8}})}^{2} - 2 = 2207$ $x^{8} + \frac{1}{x^{8}} = \sqrt{2209} = 47$ $x^{8} + \frac{1}{x^{8}} = {(x^{4} + \frac{1}{x^{4}})}^{2} - 2 = 47$ $x^{4} + \frac{1}{x^{4}} = \sqrt{49} = 7$ $x^{4} + \frac{1}{x^{4}} = {(x^{2} + \frac{1}{x^{2}})}^{2} - 2 = 7$ $x^{2} + \frac{1}{x^{2}} = \sqrt{9} = 3$ $x^{2} + \frac{1}{x^{2}} = {(x + \frac{1}{x})}^{2} - 2 = 3$ $x + \frac{1}{x} = \sqrt{5}$ $x^{2} - x \sqrt{5} + 1 = 0$ $x = \frac{\pm 1 + \sqrt{5}}{2} = φ^{\pm 1}$ $φ^{n} = φ^{n - 2} + φ^{n - 1}$ $φ^{n} = φ^{n + 2} - φ^{n + 1}$ $S = \underset{n = 1}{\sum^{2021}} {(φ^{n} + φ^{- n})}^{2}$ $S = \underset{n = 1}{\sum^{2021}} (φ^{2 n} + 2 + φ^{- 2 n})$ $S = \frac{(φ^{2}) ({(φ^{2})}^{2021} - 1)}{φ^{2} - 1} + 2 (2021) + \frac{(φ^{- 2}) ({(φ^{- 2})}^{2021} - 1)}{φ^{- 2} - 1}$ $S = \frac{(φ^{2}) ({(φ)}^{4042} - 1)}{φ} + 4042 - \frac{(φ^{- 2}) ({(φ)}^{- 4042} - 1)}{φ^{- 1}}$ $S = φ^{4043} + 4041 - {(φ^{- 1})}^{4043}$
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