Let - and - be the roots of x2-x-1-0- with - For all positive integers n- define- a n - n - n - n - 1b 1-1 and b n - a n -1- a n -1- n - 2Then which of the following option is-are correct- A- a 1- a 2- a 3- a n-an-2-1 such that n - 1B- -1 a n -10-10 n -10-89C- -n-1 b n -10 n -8-89D- bn-n-n such that n - 1

The correct options are A a_1+ a_2+a_3+......+a_n = a_n+2 1 such that n ≥1 nbsp; B ∑ _n=1^∞a_n/10^n= 10/89 nbsp; D b_n = α^n +β^n such that n≥ 1α , β are ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XIII

Mathematics

Relation between Roots and Coefficients for Quadratic

Let α and β b...

Question

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 is/are correct?

a_{1} + a_{2} + a_{3} + . . . . . . + a_{n} = a_{n + 2} - 1 such that n \geq 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

b_{n} = α^{n} + β^{n}

such that

n \geq 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct options are
A $a_{1} + a_{2} + a_{3} + . . . . . . + a_{n} = a_{n + 2} - 1 such that n \geq 1$
B $\infty \sum n = 1 \frac{a_{n}}{10^{n}} = \frac{10}{89}$
D $b_{n} = α^{n} + β^{n}$ such that $n \geq 1$
$α, β$ are roots of equation $x^{2} - x - 1 = 0$ ,
So,
$α + β = 1, α \cdot β = - 1,$
$α, β = \frac{+ 1 \pm \sqrt{5}}{2}, α - β = \sqrt{5}$
$α^{2} = {(\frac{1 + \sqrt{5}}{2})}^{2} = \frac{3 + \sqrt{5}}{2}$
$β^{2} = {(\frac{1 - \sqrt{5}}{2})}^{2} = \frac{3 - \sqrt{5}}{2}$
L.H.S = $a_{1} + a_{2} + a_{3} + . . . . . . + a_{n}$
$= \frac{(α + α^{2} + α^{3} + . . . + α^{n}) - (β + β^{2} + β^{3} + . . . + β^{n})}{α - β}$
$= \frac{α (1 - α^{n})}{(α - β) (1 - α)} - \frac{β (1 - β^{n})}{(α - β) (1 - β)}$
$α^{2} - α - 1 = 0 \Rightarrow α^{2} - 1 = α \Rightarrow α + 1 = - \frac{α}{1 - α}$
$= \frac{(- (α + 1) (1 - α^{n})) + (1 + β) (1 - β^{n})}{α - β}$
$= \frac{- α^{2} (1 - α^{n}) + β^{2} (1 - β^{n})}{α - β} = \frac{(α^{n + 2} - β^{n + 2}) + (β^{2} - α^{2})}{α - β}$
$= \frac{α^{n + 2} - β^{n + 2}}{α - β} - \frac{α^{2} - β^{2}}{α - β} = a_{n + 2} - (α + β) = a_{n + 2} - 1$

$\infty \sum n = 1 \frac{a_{n}}{10^{n}} = \frac{α^{n} - β^{n}}{(α - β) 10^{n}} = \frac{(\frac{α}{10})^{n} - (\frac{β}{10})^{n}}{α - β}$
$\infty \sum n = 1 (\frac{α}{10})^{n} = \frac{\frac{α}{10}}{1 - \frac{α}{10}} = \frac{α}{10 - α}$
Similarly, $\infty \sum n = 1 (\frac{β}{10})^{n} = \frac{β}{10 - β}$
$\infty \sum n = 1 \frac{a_{n}}{10^{n}} = \frac{\frac{α}{10 - α} - \frac{β}{10 - β}}{α - β} = \frac{α (10 - β) - β (10 - α)}{(α - β) (10 - α) (10 - β)}$
$= \frac{10 α - 10 β}{(α - β) (10 - α) (10 - β)} = \frac{10}{(10 - α) (10 - β)}$
$\frac{10}{[10 - \frac{1 + \sqrt{5}}{2}] [10 - \frac{1 - \sqrt{5}}{2}]}$
$= \frac{10 \times 4}{(20 - 1 - \sqrt{5}) (20 - 1 + \sqrt{5})} = \frac{40}{19^{2} - (\sqrt{5})^{2}} = \frac{40}{356} = \frac{10}{89}$
$b_{n} = a_{n - 1} + a_{n + 1}$
$= \frac{α^{n - 1} - β^{n - 1}}{α - β} + \frac{α^{n + 1} - β^{n + 1}}{α - β}$
$= \frac{α^{n - 1} + α^{n + 1}}{α - β} - \frac{β^{n + 1} + β^{n - 1}}{α - β} = \frac{α^{n} (\frac{α^{2} + 1}{α}) - β^{n} (\frac{β^{2} + 1}{β})}{\sqrt{5}}$
$= \frac{α^{n} β (α^{2} + 1) - β^{n} α (β^{2} + 1)}{α β \sqrt{5}} = \frac{α^{n} \frac{1 - \sqrt{5}}{2} (1 + \frac{3 + \sqrt{5}}{2}) - β^{n} \frac{1 + \sqrt{5}}{2} (1 + \frac{3 - \sqrt{5}}{2})}{- 1 \times \sqrt{5}}$
$= \frac{α^{n} (1 - \sqrt{5}) (5 + \sqrt{5}) - β^{n} (1 + \sqrt{5}) (5 - \sqrt{5})}{- 4 \times \sqrt{5}}$
$= \frac{α^{n} (- 4 \sqrt{5}) - β^{n} (4 \sqrt{5})}{- 4 \sqrt{5}} = α^{n} + β^{n}$
$\infty \sum n = 1 \frac{b_{n}}{10^{n}} = \frac{α^{n} + β^{n}}{10^{n}} = (\frac{α}{10})^{n} + (\frac{β}{10})^{n}$
$\infty \sum n = 1 (\frac{α}{10})^{n} = \frac{\frac{α}{10}}{1 - \frac{α}{10}} = \frac{α}{10 - α}$
Similarly, $\infty \sum n = 1 (\frac{β}{10})^{n} = \frac{β}{10 - β}$
$\infty \sum n = 1 \frac{b_{n}}{10^{n}} = \frac{α}{10 - α} + \frac{β}{10 - β} = \frac{α (10 - β) + β (10 - α)}{(10 - α) (10 - β)}$

$= \frac{10 (α + β) - 2 α β}{(10 - α) (10 - β)} = \frac{(10 \times 1) - 2 (- 1)}{[10 - \frac{1 + \sqrt{5}}{2}] [10 - \frac{1 - \sqrt{5}}{2}]}$
$= \frac{(10 + 2) \times 4}{[19 - \sqrt{5}] [19 + \sqrt{5}]} = \frac{12 \times 4}{(361 - 5)} = \frac{48}{356} = \frac{12}{89}$

Suggest Corrections

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 option is/are correct?

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 is/are correct?