If $a_{n} = x^{n} + \frac{1}{x^{n}},$ n being a natural number, then which of the following is equal to $a_{n + 3}$ ?

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Solution

The correct option is A

Approach :1 Using the options

$a_{1} . a_{n + 2}$ exists in all options.
We will calculate it first.
$a_{1} . a_{n + 2}$ exists in all options.
We will calculate it first.
$a_{1} . a_{n + 2} = (x + \frac{1}{x}) (x^{n + 2} + \frac{1}{x^{n + 2}})$

$= x^{n + 3} + \frac{1}{x^{n + 3}} + x^{n + 1} + \frac{1}{x^{n + 1}}$

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

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

Approach 2: Assumption
Since the question is of the form "Variable to Variable”, assume any value for "n” and "x”
Assuming n=1 $&$ x=1 . Then $a_{1} = 2$ . Similarly $a_{2} = 2$ , $a_{3} = 2$ $& a_{4} = 2$
Eliminating answer options
Option (a) =2
Option (b) $\neq$ 2

Option (c) $\neq$ 2 $&$ option (d) $\neq$ 2 . Answer is option (a)

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