Let \( \alpha, \beta; \, \alpha > \beta \), be the roots of the equation $$ x^2 - \sqrt{2}x - \sqrt{3} = 0. $$ Let \( P_n = \alpha^n - \beta^n, \, n \in \mathbb{N} \). Then $$ \left( 11\sqrt{3} - 10\sqrt{2} \right) P_{10} + \left( 11\sqrt{2} + 10 \right) P_{11} - 11P_{12} $$ is equal to:
We are given that α and β are the roots of the quadratic equation:
\(x^2 - \sqrt{2}x - \sqrt{3} = 0\)
Step 1: Find the roots α and β
To find α and β, we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For the given equation, \(a = 1\), \(b = -\sqrt{2}\), and \(c = -\sqrt{3}\). Substituting these values into the quadratic formula:
\[ x = \frac{\sqrt{2} \pm \sqrt{(\sqrt{2})^2 - 4(1)(-\sqrt{3})}}{2(1)} \]
\[ x = \frac{\sqrt{2} \pm \sqrt{2 + 4\sqrt{3}}}{2} \]
Thus, the roots α and β are:
\[ \alpha = \frac{\sqrt{2} + \sqrt{2 + 4\sqrt{3}}}{2}, \quad \beta = \frac{\sqrt{2} - \sqrt{2 + 4\sqrt{3}}}{2} \]
Step 2: Use recurrence relation for \(P_n\)
We are given that \(P_n = \alpha^n - \beta^n\). From the given quadratic equation, we know that:
\[ \alpha + \beta = \sqrt{2}, \quad \alpha\beta = -\sqrt{3} \]
Using this, we can derive a recurrence relation for \(P_n\). The recurrence relation is:
\[ P_n = (\alpha + \beta)P_{n-1} - \alpha\beta P_{n-2} \]
Substituting the values \(\alpha + \beta = \sqrt{2}\) and \(\alpha\beta = -\sqrt{3}\), we get:
\[ P_n = \sqrt{2}P_{n-1} + \sqrt{3}P_{n-2} \]
Step 3: Calculate the required expression
Now, we need to calculate the following expression:
\[ (11\sqrt{3} - 10\sqrt{2})P_{10} + (11\sqrt{2} + 10)P_{11} - 11P_{12} \]
Using the recurrence relation for \(P_n\), we can express each term in terms of \(P_9\):
\[ P_{10} = \sqrt{2}P_9 + \sqrt{3}P_8 \]
\[ P_{11} = \sqrt{2}P_{10} + \sqrt{3}P_9 \]
\[ P_{12} = \sqrt{2}P_{11} + \sqrt{3}P_{10} \]
Substituting these into the original expression, and simplifying, we find that the value of the expression is:
\[ 10\sqrt{3}P_9 \]
Thus, the correct answer is:
\[ 10\sqrt{3}P_9 \]
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is: