To solve this problem, we first need to understand the properties of the roots \( \alpha \) and \( \beta \) of the given quadratic equation \( x^2 - x - 1 = 0 \).
The solutions to this equation can be found using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -1 \), and \( c = -1 \).
This gives us:
\(x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}\)
Thus, the roots are: \( \alpha = \frac{1 + \sqrt{5}}{2} \) and \( \beta = \frac{1 - \sqrt{5}}{2} \).
Next, we use the relationships for powers of roots of quadratic equations. Since \( \alpha \) and \( \beta \) are roots of the equation \( x^2 - x - 1 = 0 \), they satisfy:
These recursive relationships can be used to express higher powers as follows:
The sequence relations for \( S_n = 2023 \alpha^n + 2024 \beta^n \) can be used to express each term in terms of previous terms:
This can be expanded to find:
Using this relation, let's analyze the options:
Thus, the correct answer is Option 2: \(2S_{11} = S_{12} + S_{10}\), as this expression is consistent with the recursive relationships derived for \( S_n \).
Given:
\(x^2 - x - 1 = 0 \implies \alpha, \beta \text{ are roots.}\)
The relation between \(\alpha\) and \(\beta\) is:
\(\alpha^2 = \alpha + 1, \quad \beta^2 = \beta + 1.\)
The sequence \(S_n\) is defined as:
\(S_n = 2023\alpha^n + 2024\beta^n.\)
Using the recurrence relation for the roots:
\(S_{n+2} = S_{n+1} + S_n.\)
Applying this for \(n = 10\):
\(S_{12} = S_{11} + S_{10}.\)
The Correct answer is: \( 2S_{11} = S_{12} + S_{10} \)
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)