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?

(A) a1+a2+a3+.....+an=an+2-1 for all n≥1

JEE Advanced · Mathematics · 2. Quadratic Equations

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?

A $a_{1} + a_{2} + a_{3} + . . . . . + a_{n} = a_{n + 2} - 1$ for all $n \geq 1$
B $\sum_{n = 1}^{\infty} \frac{a_{n}}{10^{n}} = \frac{10}{89}$
C $\sum_{n = 1}^{\infty} \frac{b_{n}}{10^{n}} = \frac{8}{89}$
D $b_{n} = α^{n} + β^{n}$ for all $n \geq 1$

Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

Given,

α

and

β

are roots of

x^{2} - x - 1 = 0

$a_{k+2}-a_k=\frac{\left(\alpha^{k+2}-\beta^{k+2}\right)-\left(\alpha^k-\beta^k\right)}{\alpha-\beta}$$=\frac{\left(\alpha^{k+2}-\alpha^k\right)-\left(\beta^{k+2}-\beta^k\right)}{\alpha-\beta}$

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

\Rightarrow a_{k + 2} - a_{k} = a_{k + 1} \Rightarrow a_{k + 2} - a_{k + 1} = a_{k}

$\sum_{k=1}^n a_k=a_{k+2}-a_2=a_{k+2}-\frac{\alpha^2-\beta^2}{\alpha-\beta}$ $=a_{k+2}-(\alpha+\beta)$

= a_{k + 2} - 1

\Rightarrow a_{1} + a_{2} + a_{3} + . . . . . + a_{n} = a_{n + 2} - 1

Option

(B)

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

= \frac{1}{α - β} \sum_{n = 1}^{\infty} ({(\frac{α}{10})}^{n} - {(\frac{β}{10})}^{n})

= \frac{1}{α - β} [\frac{\frac{α}{10}}{1 - \frac{α}{10}} - \frac{\frac{β}{10}}{1 - \frac{β}{10}}]

= \frac{1}{α - β} [\frac{α}{10 - α} - \frac{β}{10 - β}]

= \frac{1}{α - β} [\frac{10 α - 10 β}{(10 - α) (10 - β)}]

= \frac{1}{α - β} [\frac{10 (α - β)}{100 - 10 (α + β) + α β}]

= \frac{10}{100 - 10 - 1} = \frac{10}{89}

Option

(C) :

$\sum_{n=1}^{\infty} \frac{b_n}{10^n}=\sum_{n=1}^{\infty} \frac{\alpha^n+\beta^n}{10^n}=\sum_{n=1}^{\infty}\left(\frac{\alpha}{10}\right)^n$ $+~\left(\frac{\beta}{10}\right)^n$

= \frac{\frac{α}{10}}{1 - \frac{α}{10}} + \frac{\frac{β}{10}}{1 - \frac{β}{10}}

apply sum of infinite G.P. with

(a = \frac{α}{10} a n d r = \frac{α}{10})

= \frac{10 (α + β) - 2 α β}{(10 - α) (10 - β)}

= \frac{10 (1) - 2 (- 1)}{100 + α β - 10 (α + β)} = \frac{12}{100 - 1 - 10} = \frac{12}{89}

Option

(D) :

As given

b_{n} = a_{n - 1} + a_{n + 1}

b_{n} = \frac{α^{n - 1} - β^{n - 1}}{α - β} + \frac{α^{n + 1} - β^{n + 1}}{α - β}

∵ α - β = 1

b_{n} = (α^{n - 1} - β^{n - 1}) + (α^{n + 1} - β^{n + 1})

Now, as

α β = 1

∴ α^{n} β = - α^{n - 1}

∴ α β^{n} = - β^{n - 1}

b_{n} = α^{n} (α - β) + β^{n} (α - β)

b_{n} = α^{n} + β^{n}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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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≥2.Then which of the following options is/are correct?

Step-by-step Solution

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?