Let α and β be the roots of x2−x−1=0 with α>β . For all positive integers n, define an=αn−βnα−β, n≥1, b1=1 and bn=an−1+an+1,n≥2 Then which of the following options is/are correct?

Given α,β are the roots of the equation x2−x−1=0hence, α=1+52, β=1−52Given an=αn−βnα−β, n≥1Given, b1=1 and bn=an−1+an+1=αn−1−βn−1α−β+αn+1−βn+1α−β=αn−1α2+1−βn−1β2+1α−β=αn−1α+2−βn−1β+2α−β=αn−15+52−βn−15−525=αn−1α−βn−1−β=αn+βnand ∑n=1∞bn10n=∑n=1∞αn+βn10n=∑n=1∞α10n+∑n=1∞β10n=α101-α10+β101-β10=α10-α+β10-β=10(α+β)-2αβ100-10(α+β)+αβ=10(1)-2-1100-10+(-1)=1289∑n=1∞an10n=∑n=1∞αn-βn(α-β)10n=15∑n=1∞α10n-∑n=1∞β10n=15α10-α-β10-β=1510(α-β)100-10(α+β)+αβ=1510×589=1089 Also a1+a2+a3+…….+an=1α-β(α-β)+α2-β2+α3-β+…….+αn-βn=1α-βα+α2+…..+αn-β+β2+……βn=1α-β1+α+α2……+αn-1+β+β2…….βn=1α-βαn+1-1α-1-βn+1-1β-1=αn+2-βn+2α-β-1=an+2-1

Let α and β be the roots of x2−x−1=0 with α>β . For all positive integers n, define an=αn−βnα−β, n≥1, b1=1 and bn=an−1+an+1,n≥2 Then which of the following options is/are correct?

Given α,β are the roots of the equation x2−x−1=0hence, α=1+52, β=1−52Given an=αn−βnα−β, n≥1Given, b1=1 and bn=an−1+an+1=αn−1−βn−1α−β+αn+1−βn+1α−β=αn−1α2+1−βn−1β2+1α−β=αn−1α+2−βn−1β+2α−β=αn−15+52−βn−15−525=αn−1α−βn−1−β=αn+βnand ∑n=1∞bn10n=∑n=1∞αn+βn10n=∑n=1∞α10n+∑n=1∞β10n=α101-α10+β101-β10=α10-α+β10-β=10(α+β)-2αβ100-10(α+β)+αβ=10(1)-2-1100-10+(-1)=1289∑n=1∞an10n=∑n=1∞αn-βn(α-β)10n=15∑n=1∞α10n-∑n=1∞β10n=15α10-α-β10-β=1510(α-β)100-10(α+β)+αβ=1510×589=1089 Also a1+a2+a3+…….+an=1α-β(α-β)+α2-β2+α3-β+…….+αn-βn=1α-βα+α2+…..+αn-β+β2+……βn=1α-β1+α+α2……+αn-1+β+β2…….βn=1α-βαn+1-1α-1-βn+1-1β-1=αn+2-βn+2α-β-1=an+2-1

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Let $α$ and $β$ be the roots of $x^{2} - x - 1 = 0$ with $α > β$ . For all positive integers n, define $a_{n} = \frac{α^{n} - β^{n}}{α - β}, n \geq 1$ , $b_{1} = 1$ and $b_{n} = a_{n - 1} + a_{n + 1}, n \geq 2$ Then which of the following options is/are correct?

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$\sum_{n = 1}^{\infty} \frac{b_{n}}{10^{n}} = \frac{8}{89}$

$b_{n} = α^{n} + β^{n}$

$a_{1} + a_{2} + a_{3} + ...... a_{n} = a_{n + 2} - 1$

$\sum_{n = 1}^{\infty} \frac{a_{n}}{10^{n}} = \frac{10}{89}$

answer is A, B, D.

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Detailed Solution

$\begin{array}{l} Given α, β are the roots of the equation x^{2} - x - 1 = 0 \\ hence, α = \frac{1 + \sqrt{5}}{2}, β = \frac{1 - \sqrt{5}}{2} \end{array}$

$\begin{matrix} Given a_{n} = \frac{α^{n} - β^{n}}{α - β}, n \geq 1 \\ G i v e n, b_{1} = 1 and \\ b_{n} = a_{n - 1} + a_{n + 1} \\ = \frac{α^{n - 1} - β^{n - 1}}{α - β} + \frac{α^{n + 1} - β^{n + 1}}{α - β} \\ = \frac{α^{n - 1} (α^{2} + 1) - β^{n - 1} (β^{2} + 1)}{α - β} \\ = \frac{α^{n - 1} (α + 2) - β^{n - 1} (β + 2)}{α - β} \\ = \frac{α^{n - 1} (\frac{5 + \sqrt{5}}{2}) - β^{n - 1} (\frac{5 - \sqrt{5}}{2})}{\sqrt{5}} \\ = α^{n - 1} (α) - β^{n - 1} (- β) = α^{n} + β^{n} \\ and \sum_{n = 1}^{\infty} \frac{b_{n}}{10^{n}} = \sum_{n = 1}^{\infty} \frac{α^{n} + β^{n}}{10^{n}} = \sum_{n = 1}^{\infty} {(\frac{α}{10})}^{n} + \sum_{n = 1}^{\infty} {(\frac{β}{10})}^{n} \end{matrix}$

$\begin{array}{l} = \frac{\frac{α}{10}}{1 - \frac{α}{10}} + \frac{\frac{β}{10}}{1 - \frac{β}{10}} = \frac{α}{10 - α} + \frac{β}{10 - β} = \frac{10 (α + β) - 2 α β}{100 - 10 (α + β) + α β} = \frac{10 (1) - 2 (- 1)}{100 - 10 + (- 1)} = \frac{12}{89} \\ \sum_{n = 1}^{\infty} \frac{a_{n}}{10^{n}} = \sum_{n = 1}^{\infty} \frac{α^{n} - β^{n}}{(α - β) 10^{n}} = \frac{1}{\sqrt{5}} (\sum_{n = 1}^{\infty} {(\frac{α}{10})}^{n} - \sum_{n = 1}^{\infty} {(\frac{β}{10})}^{n}) = \frac{1}{\sqrt{5}} (\frac{α}{10 - α} - \frac{β}{10 - β}) \\ = \frac{1}{\sqrt{5}} (\frac{10 (α - β)}{100 - 10 (α + β) + α β}) = \frac{1}{\sqrt{5}} (\frac{10 \times \sqrt{5}}{89}) = \frac{10}{89} Also a_{1} + a_{2} + a_{3} + \dots \dots . + a_{n} \\ = \frac{1}{α - β} [(α - β) + (α^{2} - β^{2}) + (α^{3} - β) + \dots \dots . + (α^{n} - β^{n})] \\ = \frac{1}{α - β} [(α + α^{2} + \dots . . + α^{n}) - (β + β^{2} + \dots \dots β^{n})] \\ = \frac{1}{α - β} [(1 + α + α^{2} \dots \dots + α^{n}) - (1 + β + β^{2} \dots \dots . β^{n})] \\ = \frac{1}{α - β} [\frac{α^{n + 1} - 1}{α - 1} - \frac{β^{n + 1} - 1}{β - 1}] = \frac{α^{n + 2} - β^{n + 2}}{α - β} - 1 = a_{n + 2} - 1 \end{array}$