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

a 1 + a 2 + a 3 + … … . + a n = a n + 2 - 1 for all n ≥ 1 , ∑ n = 1 ∞ a n 10 n = 10 89 , ∑ n = 1 ∞ b n 10 n = 12 89 , b n = α n + β n for all n ≥ 1

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

α , β are roots of x 2 - x - 1 a r + 2 − a r = α r + 2 − β r + 2 − α r − β r α − β = α r + 2 − α r − β r + 2 − β r α − β = α r α 2 − 1 − β r β 2 − 1 α − β = α r α − β r β α − β = α r + 1 − β r + 1 α − β = a r + 1 ⇒ a r + 2 − a r + 1 = a r ⇒ ∑ r = 1 n a r = a n + 2 − a 2 = a n + 2 − α 2 − β 2 α − β = a n + 2 − α + β = a n + 2 − 1 now, ∑ n = 1 ∞ a n 10 n = ∑ N = 1 ∞ α 10 n − ∑ N = 1 ∞ β 10 n α − β α 10 1 − α 10 − β 10 1 − β 10 α − β = α 10 − α − β 10 − β α − β = 10 10 − α 10 − β = 10 89 ∑ n = 1 ∞ b n 10 n = ∑ n = 1 ∞ a n − 1 + a n + 1 10 n = α 10 1 − α 10 + β 10 1 − β 10 = 12 89 further b n = a n − 1 + a n + 1 = α n − 1 − β n − 1 + α n + 1 − β n + 1 α − β α β = − 1 ⇒ α n − 1 = α n β & β n − 1 = − α β n = α n α − β + α − β β n α − β = α n + β n

a 1 + a 2 + a 3 + … … . + a n = a n + 2 - 1 for all n ≥ 1

b n = α n + β n for all n ≥ 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 option(s) is/are correct?

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$a_{1} + a_{2} + a_{3} + \dots \dots . + a_{n} = a_{n + 2} - 1 for all n \geq 1$

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

$\sum_{n = 1}^{\infty} \frac{b_{n}}{10^{n}} = \frac{12}{89}$

$b_{n} = α^{n} + β^{n} for all n \geq 1$

answer is A, B, C, D.

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Detailed Solution

$α, β are roots of x^{2} - x - 1$

$a_{r + 2} - a_{r} = \frac{(α^{r + 2} - β^{r + 2}) - (α^{r} - β^{r})}{α - β} = \frac{(α^{r + 2} - α^{r}) - (β^{r + 2} - β^{r})}{α - β}$

$= \frac{α^{r} (α^{2} - 1) - β^{r} (β^{2} - 1)}{α - β} = \frac{α^{r} α - β^{r} β}{α - β} = \frac{α^{r + 1} - β^{r + 1}}{α - β} = a_{r + 1}$

$\Rightarrow a_{r + 2} - a_{r + 1} = a_{r}$

$\Rightarrow \sum_{r = 1}^{n} a_{r} = a_{n + 2} - a_{2} = a_{n + 2} - \frac{α^{2} - β^{2}}{α - β} = a_{n + 2} - (α + β) = a_{n + 2} - 1$

now, $\sum_{n = 1}^{\infty} \frac{a_{n}}{10^{n}} = \frac{\sum_{N = 1}^{\infty} {(\frac{α}{10})}^{n} - \sum_{N = 1}^{\infty} {(\frac{β}{10})}^{n}}{α - β}$

$\frac{\frac{\frac{α}{10}}{1 - \frac{α}{10}} - \frac{\frac{β}{10}}{1 - \frac{β}{10}}}{α - β} = \frac{\frac{α}{10 - α} - \frac{β}{10 - β}}{(α - β)} = \frac{10}{(10 - α) (10 - β)} = \frac{10}{89}$

$\sum_{n = 1}^{\infty} \frac{b_{n}}{10^{n}} = \sum_{n = 1}^{\infty} \frac{a_{n - 1} + a_{n + 1}}{10^{n}} = \frac{\frac{α}{10}}{1 - \frac{α}{10}} + \frac{\frac{β}{10}}{1 - \frac{β}{10}} = \frac{12}{89}$