Let be positive root of the equation x^2-x-1=0, and set a_n=1 5 (… - Vidyadip

MCQ

Let $\lambda$ be positive root of the equation $x^2-x-1=0$, and set $a_n=\frac{1}{\sqrt{5}}\left(\lambda^n-(1-\lambda)^n\right)$ for $n \in N$, where $N$ is the set of all natural numbers. Consider the sets $A =\left\{ n \in N : a _{ n }\right.$ is a rational number, but not an integer$\}$, and $B =\left\{ n \in N : a _{ n }\right.$ is a irrational number$\}$ Then

✓
both the sets $A$ and $B$ are empty
B
the set $A$ is empty but the set $B$ is non-empty
C
the set $A$ is non-empty and the set $B$ is empty
D
both the sets $A$ and $B$ are non-empty

✓

Answer

Correct option: A.

both the sets $A$ and $B$ are empty

a
(a)

$x ^2- x -1=0$

$x =\frac{1 \pm \sqrt{5}}{2}$

$\lambda=\frac{1+\sqrt{5}}{2}, 1-\lambda=\frac{1-\sqrt{5}}{2}$

$a _{ n }=\frac{1}{\sqrt{5}}\left(\lambda^{ n }-(1-\lambda)^{ n }\right) \forall n \in N$

$a _{ n }=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{ n }-\left(\frac{1-\sqrt{5}}{2}\right)^{ n }\right)$

$a _{ n }=\frac{1}{\sqrt{5} \cdot 2^{ n }}\left[(1+\sqrt{5})^{ n }-(1-\sqrt{5})^{ n }\right]$

$(1+\sqrt{5})^{ n }=1+{ }^{ n } C_1 \sqrt{5}+{ }^{ n } C_2(\sqrt{5})^2+\ldots$

$(1-\sqrt{5})^{ n }=1-{ }^{ n } C_1 \sqrt{5}+{ }^{ n } C_2(\sqrt{5})^2-\ldots$

$(1+\sqrt{5})^{ n }-(1-\sqrt{5})^{ n }$

$=2\left[{ }^{ n } C _1 \sqrt{5}+{ }^{ n } C_3 5 \sqrt{5}+{ }^{ n } C_5 5^2 \sqrt{5}+\ldots\right]$

So $a _{ n }=\frac{\left[{ }^{ n } C_1+{ }^{ n } C_3 5+{ }^{ n } C_5 5^2+\ldots\right]}{2^{ n -1}}$

$a _{ n }=\frac{2 \Sigma^{ n } C_{ r } 5^{ r }}{2^{ n }}$ where $r$ is odd

$a _{ n }=\frac{(1+5)^{ n }-(1-5)^{ n }}{2^{ n }}=\frac{6^{ n }-(-4)^{ n }}{2^{ n }}$

$a _{ n }=3^{ n }-(-2)^{ n } \in I \forall n \in N$

$\Rightarrow A \in \phi ; B \in \phi$

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