Let lambda be the positive root of the equati | vedclass

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(A) The given equation is $x^2-x-1=0$. The roots are $x = \frac{1 \pm \sqrt{5}}{2}$.Given $\lambda = \frac{1+\sqrt{5}}{2}$,then $1-\lambda = \frac{1-\

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

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(A) The given equation is $x^2-x-1=0$. The roots are $x = \frac{1 \pm \sqrt{5}}{2}$.Given $\lambda = \frac{1+\sqrt{5}}{2}$,then $1-\lambda = \frac{1-\sqrt{5}}{2}$.Note that $a_n$ is the $n$-th Fibonacci number $F_n$,defined by the Binet formula.$a_n = \frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}ight)^n - \left(\frac{1-\sqrt{5}}{2}ight)^night)$.Expanding using the binomial theorem:$(1+\sqrt{5})^n = \sum_{k=0}^n \binom{n}{k} (\sqrt{5})^k$ and $(1-\sqrt{5})^n = \sum_{k=0}^n \binom{n}{k} (-\sqrt{5})^k$.Then $(1+\sqrt{5})^n - (1-\sqrt{5})^n = 2 \sum_{k 	ext{ odd}} \binom{n}{k} 5^{(k-1)/2} \cdot 5^{1/2} = 2\sqrt{5} \sum_{j=0}^{\lfloor (n-1)/2 floor} \binom{n}{2j+1} 5^j$.Thus,$a_n = \frac{1}{\sqrt{5}} \cdot \frac{1}{2^n} \cdot 2\sqrt{5} \sum_{j=0}^{\lfloor (n-1)/2 floor} \binom{n}{2j+1} 5^j = \frac{1}{2^{n-1}} \sum_{j=0}^{\lfloor (n-1)/2 floor} \binom{n}{2j+1} 5^j$.Since $a_n$ is the $n$-th Fibonacci number,$a_n$ is an integer for all $n \in N$.Therefore,$a_n$ is never a non-integer rational number,so $A = \emptyset$.Also,$a_n$ is never an irrational number,so $B = \emptyset$.Hence,both sets $A$ and $B$ are empty.

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