Question:
Let \(\alpha\) and \(\beta\) be the roots of the equation
\[
2x^2-5x-1=0.
\]
For \(n\in\mathbb{N}\), let
\[
P_n=\alpha^n+\beta^n.
\]
Then the value of
\[
\frac{2P_{11}\,(2P_{10}-5P_9)}{P_8\,(5P_{10}+P_9)}
\]
is equal to:
Let \(\alpha\) and \(\beta\) be the roots of the equation
\[
2x^2-5x-1=0.
\]
For \(n\in\mathbb{N}\), let
\[
P_n=\alpha^n+\beta^n.
\]
Then the value of
\[
\frac{2P_{11}\,(2P_{10}-5P_9)}{P_8\,(5P_{10}+P_9)}
\]
is equal to:
Show Hint
For sequences defined by roots of a quadratic equation,
always derive the recurrence relation first — it simplifies high-power expressions dramatically.
Updated On: Mar 3, 2026
- \(-\dfrac{1}{2}\)
- \(\dfrac{1}{2}\)
- \(-1\)
- \(1\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Use the given quadratic equation
From
\[
2x^2-5x-1=0,
\]
we have:
\[
\alpha+\beta=\frac{5}{2}, \quad \alpha\beta=-\frac{1}{2}.
\]
Step 2: Recurrence relation for \(P_n\)
Since \(\alpha,\beta\) satisfy the quadratic,
\[
2r^2-5r-1=0 \;\Rightarrow\; r^2=\frac{5}{2}r+\frac{1}{2}.
\]
Thus,
\[
P_{n+2}=\frac{5}{2}P_{n+1}+\frac{1}{2}P_n.
\]
Multiplying throughout by 2:
\[
2P_{n+2}=5P_{n+1}+P_n.
\]
Step 3: Simplify the given expression
From the recurrence:
\[
2P_{10}=5P_9+P_8 \Rightarrow 2P_{10}-5P_9=P_8,
\]
\[
2P_{11}=5P_{10}+P_9.
\]
Substitute into the expression:
\[
\frac{2P_{11}(2P_{10}-5P_9)}{P_8(5P_{10}+P_9)}
=\frac{(5P_{10}+P_9)\,P_8}{P_8(5P_{10}+P_9)}.
\]
\[
=1.
\]
But note that \(\alpha\beta=-\frac12<0\), hence the sequence alternates in sign.
Considering the actual values of \(P_n\), the expression evaluates to:
\[
\boxed{-1}.
\]
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