JEE Challenger
More from Complex Numbers and Quadratic Equations

Evaluate Twelfth Term in Sequence Defined by Quadratic Roots

Comprehension Passage

Suppose p,qp, q are integers and let α,β\alpha, \beta be the distinct roots of x2−x−1=0x^2 - x - 1 = 0. For n=0,1,2,…n = 0, 1, 2, \dots, let an=pαn+qβna_n = p \alpha^n + q \beta^n.

FACT: If aa and bb are rational numbers satisfying a+b5=0a + b\sqrt{5} = 0, then a=0=ba = 0 = b.

Determine the value of a12a_{12}:

Options

A

a11−a10a_{11} - a_{10}

B

a11+a10a_{11} + a_{10}

Correct
C

2a11+a102a_{11} + a_{10}

D

a11+2a10a_{11} + 2a_{10}

Step-by-Step Solution

Given that α\alpha and β\beta are the distinct roots of the quadratic equation: x2−x−1=0x^2 - x - 1 = 0

Since α\alpha and β\beta satisfy this equation, we have: α2−α−1=0  ⟹  α2=α+1\alpha^2 - \alpha - 1 = 0 \implies \alpha^2 = \alpha + 1 β2−β−1=0  ⟹  β2=β+1\beta^2 - \beta - 1 = 0 \implies \beta^2 = \beta + 1

Multiplying both sides of the equations by αn\alpha^n and βn\beta^n respectively for any non-negative integer nn, we get: αn+2=αn+1+αn\alpha^{n+2} = \alpha^{n+1} + \alpha^n βn+2=βn+1+βn\beta^{n+2} = \beta^{n+1} + \beta^n

Now, multiplying the first equation by pp and the second equation by qq, and adding them together yields: pαn+2+qβn+2=(pαn+1+qβn+1)+(pαn+qβn)p \alpha^{n+2} + q \beta^{n+2} = (p \alpha^{n+1} + q \beta^{n+1}) + (p \alpha^n + q \beta^n)

Using the definition an=pαn+qβna_n = p \alpha^n + q \beta^n, this recurrence relation simplifies to: an+2=an+1+ana_{n+2} = a_{n+1} + a_n

To find a12a_{12}, we substitute n=10n = 10 into the recurrence relation: a12=a11+a10a_{12} = a_{11} + a_{10}

Thus, the correct option is B.