Question

If $\left(x^2 + \frac{1}{x^2}\right) = 25$ and $x>0$, then the value of $\left(x^7 + \frac{1}{x^7}\right)$ is

• A. \(44859\sqrt{3}\)
• B. \(44853\sqrt{3}\)
• C. \(44850\sqrt{3}\)
• D. \(44856\sqrt{3}\)

Hint:

For expressions like $x^n + \dfrac{1}{x^n}$, first find $x + \dfrac{1}{x}$ using the given data, then use the recurrence \[ S_n = S_1 S_{n-1} - S_{n-2}, \] instead of expanding powers directly. It saves a lot of time and reduces mistakes in exams.

Answer 1

The Correct Option is B

Step 1: Find \(x + \frac{1}{x}\). We are given: \[ x^2 + \frac{1}{x^2} = 25. \] Recall: \[ \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2}. \] So, \[ \left(x + \frac{1}{x}\right)^2 = 25 + 2 = 27 \quad \Rightarrow \quad x + \frac{1}{x} = \sqrt{27} = 3\sqrt{3}, \] since \(x > 0\) implies \(x + \frac{1}{x} > 0\). Let \[ S_1 = x + \frac{1}{x} = 3\sqrt{3}. \]  
Step 2: Use identities / recurrence to build higher powers. We use the standard recurrence: \[ S_n = x^n + \frac{1}{x^n}, \quad \text{and} \quad S_n = S_1 S_{n-1} - S_{n-2}, \text{ for } n \ge 2. \] We already know: \[ S_1 = 3\sqrt{3}, \quad S_2 = x^2 + \frac{1}{x^2} = 25. \] Now compute step by step. 
\(\underline{For }\) \(n = 3\): \[ S_3 = S_1 S_2 - S_1 = (3\sqrt{3})(25) - 3\sqrt{3} = 72\sqrt{3}. \] \(\underline{For } \)\(n = 4\): \[ S_4 = S_1 S_3 - S_2 = (3\sqrt{3})(72\sqrt{3}) - 25 = 3 \cdot 72 \cdot 3 - 25 = 648 - 25 = 623. \]\( \underline{For }\) \(n = 5\): \[ S_5 = S_1 S_4 - S_3 = (3\sqrt{3})(623) - 72\sqrt{3} = (1869 - 72)\sqrt{3} = 1797\sqrt{3}. \] \(\underline{For }\) \(n = 6\): \[ S_6 = S_1 S_5 - S_4 = (3\sqrt{3})(1797\sqrt{3}) - 623 = 3 \cdot 1797 \cdot 3 - 623 = 16173 - 623 = 15550. \] \(\underline{For }\) \(n = 7\): \[ S_7 = S_1 S_6 - S_5 = (3\sqrt{3})(15550) - 1797\sqrt{3} = (46650 - 1797)\sqrt{3} = 44853\sqrt{3}. \] Thus, \[ x^7 + \frac{1}{x^7} = 44853\sqrt{3}. \]