a + (1/a) = 3 find: a^5 + (1/a^5 ) = ?


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Question Number 199135 by hardmath last updated on 28/Oct/23

$a + \frac{1}{a} = 3$ $find : a^{5} + \frac{1}{a^{5}} = ?$
	Answered by som(math1967) last updated on 28/Oct/23

	$123$
	Answered by a.lgnaoui last updated on 28/Oct/23

	$3^{5} = a^{5} + \frac{1}{a^{5}} + 5 (a^{3} + \frac{1}{a^{3}}) + 10 (a + \frac{1}{a})$ $= a^{5} + \frac{1}{a^{5}} + 5 (a + \frac{1}{a}) [a^{2} + \frac{1}{a^{2}} + 1] =$ $= a^{5} + \frac{1}{a^{5}} + 5 (a + \frac{1}{a}) [{(a + \frac{1}{a})}^{2} - 1]$ $\Rightarrow a^{5} + \frac{1}{a^{5}} = 3^{5} - 5 \times 24 = 243 - 120$ $a^{5} + \frac{1}{a^{5}} = 123$
	Commented by hardmath last updated on 28/Oct/23

	$thank you professor$
	Answered by Rasheed.Sindhi last updated on 28/Oct/23

	$a + \frac{1}{a} = 3; a^{5} + \frac{1}{a^{5}} = ?$ ${(a + \frac{1}{a})}^{2} = 3^{2}$ $a^{2} + \frac{1}{a^{2}} = 9 - 2 = 7$ ${(a + \frac{1}{a})}^{3} = 3^{3}$ $a^{3} + \frac{1}{a^{3}} = 27 - 3 (a + \frac{1}{a}) = 27 - 3 (3) = 18$ $(a^{2} + \frac{1}{a^{2}}) (a^{3} + \frac{1}{a^{3}}) = a^{5} + \frac{1}{a^{5}} + a + \frac{1}{a}$ $(7) (18) = a^{5} + \frac{1}{a^{5}} + 3$ $a^{5} + \frac{1}{a^{5}} = 7 \times 18 - 3 = 123$
	Answered by Rasheed.Sindhi last updated on 28/Oct/23

	$a + \frac{1}{a} = 3; a^{5} + \frac{1}{a^{5}} = ?$ $∙ {(a + \frac{1}{a})}^{2} = 3^{2} = 9$ $a^{2} + \frac{1}{a^{2}} = 9 - 2 = 7$ $∙ {(a^{2} + \frac{1}{a^{2}})}^{2} = 7^{2} = 49$ $a^{4} + \frac{1}{a^{4}} = 49 - 2 = 47$ $∙ (a^{2} + \frac{1}{a^{2}}) (a + \frac{1}{a}) = (7) (3) = 21$ $a^{3} + \frac{1}{a^{3}} + a + \frac{1}{a} = 21$ $a^{3} + \frac{1}{a^{3}} = 21 - (a + \frac{1}{a}) = 21 - 3 = 18$ $(a^{4} + \frac{1}{a^{4}}) (a + \frac{1}{a}) = (47) (3)$ $a^{5} + \frac{1}{a^{5}} + a^{3} + \frac{1}{a^{3}} = 141$ $a^{5} + \frac{1}{a^{5}} + 18 = 141$ $a^{5} + \frac{1}{a^{5}} = 141 - 18 = 123$
	Commented by hardmath last updated on 28/Oct/23

	$thank you professor$
	Answered by ajfour last updated on 28/Oct/23
	$a^2 +(1/a^2 )=7 p^n +q^n =(p+q){p^(n−1) −qp^(n−2) +.... ....+p(−q)^(n−2) +(−q)^(n−1) } hence a^5 +(1/a^5 )=(a+(1/a))(a^4 −a^2 +1−(1/a^2 )+(1/a^4 )) ⇒ Q=3(a^4 +(1/a^4 )−a^2 −(1/a^2 )+1) =3(47−7+1) = 3×41=123$
	$a^{2} + \frac{1}{a^{2}} = 7$ $p^{n} + q^{n} = (p + q) {p^{n - 1} - {q p}^{n - 2} + . . . .$ $. . . . + p {(- q)}^{n - 2} + {(- q)}^{n - 1}}$ $h e n c e$ $a^{5} + \frac{1}{a^{5}} = (a + \frac{1}{a}) (a^{4} - a^{2} + 1 - \frac{1}{a^{2}} + \frac{1}{a^{4}})$ $\Rightarrow Q = 3 (a^{4} + \frac{1}{a^{4}} - a^{2} - \frac{1}{a^{2}} + 1)$ $= 3 (47 - 7 + 1)$ $= 3 \times 41 = 123$
	Answered by Rasheed.Sindhi last updated on 28/Oct/23
	$a + (1/a) = 3; a^5 + (1/a^5 ) = ? a + (1/a) = 3⇒ { (((1/a)=3−a^★ )),((a^2 =3a−1^(★★) )) :} ^★ (1/a)=3−a (1/a^2 )=(3−a)^2 =a^2 −6a+9 =(3a−1)−6a+9=−3a+8 (1/a^3 )=((−3a+8)/a)=−3+8((1/a)) =−3+8(3−a)=−8a+21 (1/a^4 )=−8+((21)/a)=−8+21(3−a) =−21a+55 (1/a^5 )=−21+((55)/a)=−21+55(3−a) =−21+165−55a=−55a+144 ^(★★) a^2 =3a−1 a^3 =3a^2 −a=3(3a−1)−a=8a−3 a^4 =8a^2 −3a=8(3a−1)−3a=21a−8 a^5 =21a^2 −8a=21(3a−1)−8a=55a−21 a^5 + (1/a^5 ) =(55a−21)+(−55a+144) =123$
	$a + \frac{1}{a} = 3; a^{5} + \frac{1}{a^{5}} = ?$ $a + \frac{1}{a} = 3 \Rightarrow {\begin{cases} \frac{1}{a} = 3 - a^{★} \\ a^{2} = 3 a - 1^{★ ★} \end{cases}$ $^{★} \frac{1}{a} = 3 - a$ $\frac{1}{a^{2}} = {(3 - a)}^{2} = a^{2} - 6 a + 9$ $= (3 a - 1) - 6 a + 9 = - 3 a + 8$ $\frac{1}{a^{3}} = \frac{- 3 a + 8}{a} = - 3 + 8 (\frac{1}{a})$ $= - 3 + 8 (3 - a) = - 8 a + 21$ $\frac{1}{a^{4}} = - 8 + \frac{21}{a} = - 8 + 21 (3 - a)$ $= - 21 a + 55$ $\frac{1}{a^{5}} = - 21 + \frac{55}{a} = - 21 + 55 (3 - a)$ $= - 21 + 165 - 55 a = - 55 a + 144$ $^{★ ★} a^{2} = 3 a - 1$ $a^{3} = {3 a}^{2} - a = 3 (3 a - 1) - a = 8 a - 3$ $a^{4} = {8 a}^{2} - 3 a = 8 (3 a - 1) - 3 a = 21 a - 8$ $a^{5} = {21 a}^{2} - 8 a = 21 (3 a - 1) - 8 a = 55 a - 21$ $a^{5} + \frac{1}{a^{5}} = (55 a - 21) + (- 55 a + 144)$ $= 123$
	Answered by Skabetix last updated on 28/Oct/23

	$a^{2} - 3 a + 1 = 0$ $△ = b^{2} - 4 a c = 9 - 4 = 5 > 0$ $x_{1} a n d x_{2} = \frac{3 \pm \sqrt{} 5}{2}$ $a^{5} + \frac{1}{a^{5}} = \frac{3 \pm \sqrt{5}}{2} + \frac{1}{\frac{3 \pm \sqrt{5}}{2}} = 123$
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