Question

If \(\alpha, \beta\) are roots of the equation \(x^2 + 5(\sqrt{2})x + 10 = 0, \alpha>\beta\) and \(P_n = \alpha^n - \beta^n\) for each positive integer n, then the value of \(\frac{P_{17}P_{20} + 5\sqrt{2} P_{17}P_{19}}{P_{18}P_{19} + 5\sqrt{2} P_{18}^2}\) is equal to __________.

Hint:

Whenever you see a ratio involving powers of roots of a quadratic, always look for the recurrence relation $aP_n + bP_{n-1} + cP_{n-2} = 0$ before trying to find the roots explicitly.

Answer 1

Correct Answer: 1

Step 1: Understanding the Concept:
Since \(\alpha\) and \(\beta\) are roots of the quadratic equation \(ax^2 + bx + c = 0\), they satisfy the equation.
A sequence defined by \(P_n = \alpha^n \pm \beta^n\) satisfies the linear recurrence relation \(a P_n + b P_{n-1} + c P_{n-2} = 0\).
Step 2: Key Formula or Approach:
Newton's sums: For \(x^2 + 5\sqrt{2}x + 10 = 0\), the relation is:
\[ P_n + 5\sqrt{2} P_{n-1} + 10 P_{n-2} = 0 \]
Step 3: Detailed Explanation:
The expression given is:
\[ E = \frac{P_{17} (P_{20} + 5\sqrt{2} P_{19})}{P_{18} (P_{19} + 5\sqrt{2} P_{18})} \]
Using the recurrence relation for \(n=20\):
\(P_{20} + 5\sqrt{2} P_{19} + 10 P_{18} = 0 \implies P_{20} + 5\sqrt{2} P_{19} = -10 P_{18}\).
Using the recurrence relation for \(n=19\):
\(P_{19} + 5\sqrt{2} P_{18} + 10 P_{17} = 0 \implies P_{19} + 5\sqrt{2} P_{18} = -10 P_{17}\).
Substitute these into the expression:
\[ E = \frac{P_{17} (-10 P_{18})}{P_{18} (-10 P_{17})} \]
The terms \(P_{17}, P_{18},\) and \(-10\) cancel out:
\[ E = 1 \]
Step 4: Final Answer:
The value of the expression is 1.