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

α, β are the roots of x² – x – 4 = 0 and
\(P_n=α^n–β^n,\)
\(\therefore I = \frac{(P_{15} - P_{14})P_{16} - P_{15}(P_{15} - P_{14})}{P_{13}P_{14}}\)
\(I = \frac{(P_{16} - P_{15})(P_{15} - P_{14})}{P_{13}P_{14}}\)
\(⇒\)\(I = \frac{(\alpha^{16} - \beta^{16} - \alpha^{15} + \beta^{15})(\alpha^{15} - \beta^{15} - \alpha^{14} + \beta^{14})}{(\alpha^{13} - \beta^{13})(\alpha^{14} - \beta^{14})}\)
\(⇒\)\(I = \frac{\alpha^{15}(\alpha - 1) - \beta^{15}(\beta - 1))(\alpha^{14}(\alpha - 1) - \beta^{14}(\beta - 1))}{(\alpha^{13} - \beta^{13})(\alpha^{14} - \beta^{14})}\)
As \(α^2–α=4\)
\(⇒\)\(α−1=\frac{4}{α}\) and \(β−1=\frac{4}{β}\)
\(⇒\)\(I = \frac{(\alpha^{15} \cdot \frac{4}{\alpha} - \beta^{15} \cdot \frac{4}{\beta})(\alpha^{14} \cdot \frac{4}{\alpha} - \beta^{14} \cdot \frac{4}{\beta})}{(\alpha^{13} - \beta^{13})(\alpha^{14} - \beta^{14})}\)
\(I = \frac{16(\alpha^{14} - \beta^{14})(\alpha^{13} - \beta^{13})}{(\alpha^{14} - \beta^{14})(\alpha^{13} - \beta^{13})}\)
=16
So, the answer is 16.