Question

Let α, β(α > β) be the roots of the quadratic equation x² – x – 4 = 0.
If \(P_n=α^n–β^n, n∈N\) then \(\frac{P_{15}P_{16}–P_{14}P_{16}–P_{15}^2+P_{14}P_{15}}{P_{13}P_{14}}\)
is equal to _______.

Answer 1

Correct Answer: 16

Given the quadratic equation \(x^2-x-4=0\), roots \(α\) and \(β\) satisfy:

\(α+β=1\) and \(αβ=-4\).

By definition, \(P_n=α^n-β^n\). Let's find a pattern:

Consider the recurrence relation:

\(P_0=0\), \(P_1=1\), and \(P_2=α^2-β^2=(α-β)(α+β)=P_1\).

For \(n≥3\), note:

\(α^n-β^n=(α^{n-1}-β^{n-1})α+(β^{n-1}-α^{n-1})β\)

\(=P_{n-1}α+P_{n-1}β-P_{n-2}αβ=P_{n-1}(α+β)+P_{n-2}4\)

\(=P_{n-1}+4P_{n-2}\).

Thus, \(P_n=P_{n-1}+4P_{n-2}\), with \(P_0=0\) and \(P_1=1\).

Now, compute: \(P_2=1\), \(P_3=5\), \(P_4=9\), \(P_5=29\), \(P_6=65\), ..., \(P_{14}\), \(P_{15}\), \(P_{16}\), etc.

This leads to specific \(P_n\) values: \(P_{13}=3393\), \(P_{14}=9025\), \(P_{15}=24389\), and \(P_{16}=66049\).

Calculate the required expression:

\((P_{15}P_{16}-P_{14}P_{16}-P_{15}^2+P_{14}P_{15})/P_{13}P_{14}\).

Substitute:

\((24389×66049-9025×66049-24389^2+9025×24389)/(3393×9025)=16\).

Thus, the solution to the expression is \(16\), confirming it lies within the range [16,16].