(((1+(√5))/2))^(15) =((a+b(√5))/2) find a+b


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Question Number 182627 by manxsol last updated on 12/Dec/22

${(\frac{1 + \sqrt{5}}{2})}^{15} = \frac{a + b \sqrt{5}}{2}$ $f i n d a + b$
	Answered by Rasheed.Sindhi last updated on 12/Dec/22
	$(((1+(√5))/2))^(15) =((a+b(√5))/2) ; a+b=? •(((1+(√5))/2))^3 =((1+5(√5) +3(√5) (1+(√5)))/2^3 ) =((1+8(√5) +15)/8)=2+(√5) •{(((1+(√5))/2))^3 }^2 =(2+(√5) )^2 =4+5+4(√5) =9+4(√5) •({(((1+(√5))/2))^3 }^2 )^2 =(9+4(√5) )^2 =81+80+72(√5) =161+72(√5) •({(((1+(√5))/2))^3 }^2 )^2 (((1+(√5))/2))^3 (((1+(√5))/2))^(15) =(161+72(√5) )(2+(√5) ) =322+144(√5) +161(√5) +360 =682+305(√5) =((2(682+305(√5) ))/2) =((1364+610(√5))/2)=((a+b(√5))/2) a+b=1364+610=1974$
	${(\frac{1 + \sqrt{5}}{2})}^{15} = \frac{a + b \sqrt{5}}{2}; a + b = ?$ $∙ {(\frac{1 + \sqrt{5}}{2})}^{3} = \frac{1 + 5 \sqrt{5} + 3 \sqrt{5} (1 + \sqrt{5})}{2^{3}}$ $= \frac{1 + 8 \sqrt{5} + 15}{8} = 2 + \sqrt{5}$ $∙ {{(\frac{1 + \sqrt{5}}{2})}^{3}}^{2} = {(2 + \sqrt{5})}^{2} = 4 + 5 + 4 \sqrt{5} = 9 + 4 \sqrt{5}$ $∙ {({{(\frac{1 + \sqrt{5}}{2})}^{3}}^{2})}^{2} = {(9 + 4 \sqrt{5})}^{2} = 81 + 80 + 72 \sqrt{5} = 161 + 72 \sqrt{5}$ $∙ {({{(\frac{1 + \sqrt{5}}{2})}^{3}}^{2})}^{2} {(\frac{1 + \sqrt{5}}{2})}^{3}$ ${(\frac{1 + \sqrt{5}}{2})}^{15} = (161 + 72 \sqrt{5}) (2 + \sqrt{5})$ $= 322 + 144 \sqrt{5} + 161 \sqrt{5} + 360$ $= 682 + 305 \sqrt{5}$ $= \frac{2 (682 + 305 \sqrt{5})}{2}$ $= \frac{1364 + 610 \sqrt{5}}{2} = \frac{a + b \sqrt{5}}{2}$ $a + b = 1364 + 610 = 1974$
	Commented by manxsol last updated on 12/Dec/22

	$t h a n k s, S i r . R a s h e e d f o r w o r k$ $. i t i s m y g i f t$ ${(\frac{1 + \sqrt{5}}{2})}^{15} = \frac{1364 + 610 \sqrt{5}}{2}$ $s e r i e d e f i n o n a c c i$ $\begin{array}{cc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\ 1 & 1 & 2 & 3 & 5 & 8 & 13 & 21 & 34 & 55 & 89 & 144 & 233 & 377 & 610 & 987 \end{array}$ $a = 610$ $b = 377 + 987 = 1364$ $d e d u c c i o n h e c h a d e t a b l a d e$ $p o t e n c i a s d e Φ = (1 + \sqrt{5}) / 2$
	Commented by Rasheed.Sindhi last updated on 12/Dec/22

	$A nice gift!$ $T han k s a lOt sir manxsol!$
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