JEE Challenger
More from Complex Numbers and Quadratic Equations
MathMedAdv 2019 • Paper 1

Properties of Sequences Defined via Roots of Quadratic Equation

Let α\alpha and β\beta be the roots of the quadratic equation x2−x−1=0x^2 - x - 1 = 0, with α>β\alpha > \beta. For all positive integers nn, define an=αn−βnα−β,n≥1,a_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}, \quad n \ge 1, b1=1andbn=an−1+an+1,n≥2.b_1 = 1 \quad \text{and} \quad b_n = a_{n-1} + a_{n+1}, \quad n \ge 2. Then which of the following statements is/are correct?

Options

A

a1+a2+a3+⋯+an=an+2−1a_1 + a_2 + a_3 + \dots + a_n = a_{n+2} - 1 for all n≥1n \ge 1

Correct
B

∑n=1∞an10n=1089\sum_{n=1}^{\infty} \frac{a_n}{10^n} = \frac{10}{89}

Correct
C

bn=αn+βnb_n = \alpha^n + \beta^n for all n≥1n \ge 1

Correct
D

∑n=1∞bn10n=889\sum_{n=1}^{\infty} \frac{b_n}{10^n} = \frac{8}{89}

Topics & Concepts

Complex Numbers and Quadratic EquationsQuadratic EquationsSum of Series

Step-by-Step Solution

To determine which of the statements are correct, let us analyze each option step-by-step:

Given: α\alpha and β\beta are the roots of the quadratic equation x2−x−1=0x^2 - x - 1 = 0, with α>β\alpha > \beta. By Vieta's formulas, we have: α+β=1,αβ=−1\alpha + \beta = 1, \quad \alpha \beta = -1 Also, α2=α+1\alpha^2 = \alpha + 1 and β2=β+1\beta^2 = \beta + 1.

The sequence ana_n is defined as: an=αn−βnα−β,n≥1a_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}, \quad n \ge 1 This is the standard Binet formula for the Fibonacci sequence: a1=1,a2=1,a3=2,a4=3,…a_1 = 1, \quad a_2 = 1, \quad a_3 = 2, \quad a_4 = 3, \quad \dots and satisfies the recurrence relation: an+2=an+1+an  ⟹  an=an+2−an+1for all n≥1a_{n+2} = a_{n+1} + a_n \implies a_n = a_{n+2} - a_{n+1} \quad \text{for all } n \ge 1

Option (A):

Summing the sequence aka_k from k=1k = 1 to nn: ∑k=1nak=∑k=1n(ak+2−ak+1)\sum_{k=1}^n a_k = \sum_{k=1}^n (a_{k+2} - a_{k+1}) (a1+a2+⋯+an)=(a3−a2)+(a4−a3)+⋯+(an+2−an+1)(a_1 + a_2 + \dots + a_n) = (a_3 - a_2) + (a_4 - a_3) + \dots + (a_{n+2} - a_{n+1}) This is a telescoping sum, which simplifies to: ∑k=1nak=an+2−a2\sum_{k=1}^n a_k = a_{n+2} - a_2 Since a2=1a_2 = 1, we obtain: a1+a2+a3+⋯+an=an+2−1a_1 + a_2 + a_3 + \dots + a_n = a_{n+2} - 1 Thus, Option (A) is correct.

Option (C):

Given b1=1b_1 = 1 and bn=an−1+an+1b_n = a_{n-1} + a_{n+1} for n≥2n \ge 2: For n=1n=1: α1+β1=α+β=1=b1\alpha^1 + \beta^1 = \alpha + \beta = 1 = b_1 For n≥2n \ge 2: bn=αn−1−βn−1α−β+αn+1−βn+1α−β=αn(1α+α)−βn(1β+β)α−βb_n = \frac{\alpha^{n-1} - \beta^{n-1}}{\alpha - \beta} + \frac{\alpha^{n+1} - \beta^{n+1}}{\alpha - \beta} = \frac{\alpha^n\left(\frac{1}{\alpha} + \alpha\right) - \beta^n\left(\frac{1}{\beta} + \beta\right)}{\alpha - \beta} Using the relations 1α=−β\frac{1}{\alpha} = -\beta and 1β=−α\frac{1}{\beta} = -\alpha (since αβ=−1\alpha\beta = -1): 1α+α=α−β\frac{1}{\alpha} + \alpha = \alpha - \beta 1β+β=β−α=−(α−β)\frac{1}{\beta} + \beta = \beta - \alpha = -(\alpha - \beta) Substituting these into the expression for bnb_n: bn=αn(α−β)−βn(−(α−β))α−β=(α−β)(αn+βn)α−β=αn+βnb_n = \frac{\alpha^n(\alpha - \beta) - \beta^n(-(\alpha - \beta))}{\alpha - \beta} = \frac{(\alpha - \beta)(\alpha^n + \beta^n)}{\alpha - \beta} = \alpha^n + \beta^n Therefore, bn=αn+βnb_n = \alpha^n + \beta^n holds for all n≥1n \ge 1. Thus, Option (C) is correct.

Option (B):

We compute the infinite sum ∑n=1∞an10n\sum_{n=1}^{\infty} \frac{a_n}{10^n}: ∑n=1∞an10n=1α−β[∑n=1∞(α10)n−∑n=1∞(β10)n]\sum_{n=1}^{\infty} \frac{a_n}{10^n} = \frac{1}{\alpha - \beta} \left[ \sum_{n=1}^{\infty} \left(\frac{\alpha}{10}\right)^n - \sum_{n=1}^{\infty} \left(\frac{\beta}{10}\right)^n \right] Since ∣α10∣<1\left|\frac{\alpha}{10}\right| < 1 and ∣β10∣<1\left|\frac{\beta}{10}\right| < 1, using the sum formula for an infinite geometric progression: ∑n=1∞an10n=1α−β[α101−α10−β101−β10]=1α−β[α10−α−β10−β]\sum_{n=1}^{\infty} \frac{a_n}{10^n} = \frac{1}{\alpha - \beta} \left[ \frac{\frac{\alpha}{10}}{1 - \frac{\alpha}{10}} - \frac{\frac{\beta}{10}}{1 - \frac{\beta}{10}} \right] = \frac{1}{\alpha - \beta} \left[ \frac{\alpha}{10 - \alpha} - \frac{\beta}{10 - \beta} \right] =1α−β[10α−αβ−10β+αβ(10−α)(10−β)]=1α−β[10(α−β)100−10(α+β)+αβ]= \frac{1}{\alpha - \beta} \left[ \frac{10\alpha - \alpha\beta - 10\beta + \alpha\beta}{(10 - \alpha)(10 - \beta)} \right] = \frac{1}{\alpha - \beta} \left[ \frac{10(\alpha - \beta)}{100 - 10(\alpha + \beta) + \alpha\beta} \right] Substituting α+β=1\alpha + \beta = 1 and αβ=−1\alpha\beta = -1: ∑n=1∞an10n=10100−10(1)+(−1)=1089\sum_{n=1}^{\infty} \frac{a_n}{10^n} = \frac{10}{100 - 10(1) + (-1)} = \frac{10}{89} Thus, Option (B) is correct.

Option (D):

Using bn=αn+βnb_n = \alpha^n + \beta^n: ∑n=1∞bn10n=∑n=1∞(α10)n+∑n=1∞(β10)n=α101−α10+β101−β10\sum_{n=1}^{\infty} \frac{b_n}{10^n} = \sum_{n=1}^{\infty} \left(\frac{\alpha}{10}\right)^n + \sum_{n=1}^{\infty} \left(\frac{\beta}{10}\right)^n = \frac{\frac{\alpha}{10}}{1 - \frac{\alpha}{10}} + \frac{\frac{\beta}{10}}{1 - \frac{\beta}{10}} =α10−α+β10−β=10(α+β)−2αβ100−10(α+β)+αβ= \frac{\alpha}{10 - \alpha} + \frac{\beta}{10 - \beta} = \frac{10(\alpha + \beta) - 2\alpha\beta}{100 - 10(\alpha + \beta) + \alpha\beta} Substituting α+β=1\alpha + \beta = 1 and αβ=−1\alpha\beta = -1: ∑n=1∞bn10n=10(1)−2(−1)100−10(1)+(−1)=1289≠889\sum_{n=1}^{\infty} \frac{b_n}{10^n} = \frac{10(1) - 2(-1)}{100 - 10(1) + (-1)} = \frac{12}{89} \neq \frac{8}{89} Thus, Option (D) is incorrect.

Conclusion:

The correct statements are (A), (B), and (C).

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