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Question Number 184738 by mnjuly1970 last updated on 11/Jan/23

$α, β a r e r o o t s o f, x^{2} - x - 1 = 0$ $(α > β) a n d, t_{n} = \frac{α^{n} - β^{n}}{α - β}$ $(n \in N), i f, b_{1} = 1, b_{n} = t_{n - 1} + t_{n - 2}$ $(n ⩾ 2) f i n d t h e v a l u e o f$ $S = \underset{n = 1}{\sum^{\infty}} \frac{b_{n}}{10^{n}} = ?$
	Answered by mr W last updated on 11/Jan/23

	$α, β = \frac{1 \pm \sqrt{5}}{2}$ $α + 1 = α^{2}$ $β + 1 = β^{2}$ $α + β = 1$ $α β = - 1$ ${(α - β)}^{2} = {(α + β)}^{2} - 4 α β = 5$ $\Rightarrow α - β = \sqrt{5}$ $t_{n} = \frac{α^{n} - β^{n}}{α - β} = \frac{1}{\sqrt{5}} (α^{n} - β^{n})$ $b_{n} = t_{n - 1} + t_{n - 2} = \frac{1}{\sqrt{5}} (α^{n - 1} + α^{n - 2} - β^{n - 1} + β^{n - 2})$ $= \frac{1}{\sqrt{5}} (α^{n - 2} (α + 1) - β^{n - 2} (β + 1))$ $= \frac{1}{\sqrt{5}} (α^{n} - β^{n})$ $\frac{b_{n}}{10^{n}} = \frac{1}{\sqrt{5}} [{(\frac{α}{10})}^{n} - {(\frac{β}{10})}^{n}]$ $S = \underset{n = 1}{\sum^{\infty}} \frac{b_{n}}{10^{n}} = \frac{1}{\sqrt{5}} (\frac{\frac{α}{10}}{1 - \frac{α}{10}} - \frac{\frac{β}{10}}{1 - \frac{β}{10}})$ $= \frac{1}{\sqrt{5}} (\frac{α}{10 - α} - \frac{β}{10 - β})$ $= \frac{10 (α - β)}{\sqrt{5} (100 - 10 (α + β) + α β)}$ $= \frac{10}{100 - 10 - 1}$ $= \frac{10}{89}$
	Commented bymnjuly1970 last updated on 11/Jan/23

	$t h a n k y o u s o m u c h s i r W$
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