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माना द्विघात समीकरण $x ^2- x -4=0$ के मूल $\alpha, \beta(\alpha > \beta)$ हैं। यदि $P _{ n }=\alpha^{ n }-\beta^{ n }, n \in N$ है, तो $\frac{ P _{15} P _{16}- P _{14} P _{16}- P _{15}^2+ P _{14} P _{15}}{ P _{13} P _{14}}$ बराबर है $.........$.
$15$
$14$
$13$
$16$
Solution
$Pn =\alpha^{ n }-\beta^{ n } \quad x ^{2}- x -4=0$
$\frac{ P _{15} P _{16}- P _{14} P _{16}- P _{15}^{2}+ P _{14} P _{15}}{ P _{13} P _{14}}$
As $P _{ n }- P _{ n -1}=\left(\alpha^{ a }-\beta^{ n }\right)-\left(\alpha^{ n -1}-\beta^{ n -1}\right)$
$=\alpha^{ n -2}\left(\alpha^{2}-\alpha\right)-\beta^{ n -2}\left(\beta^{2}-\beta\right)$
$=4\left(\alpha^{ n -2}-\beta^{ n -2}\right)$
$P _{ a }- P _{ n -1}=4 P _{ n -2}$
Hence Expression $(1)$
$\frac{P_{16}\left(P_{15}-P_{14}\right)-P_{15}\left(P_{15}-P_{14}\right)}{P_{13} P_{14}}$
$=\frac{\left( P _{15}- P _{14}\right)\left( P _{16}- P _{15}\right)}{ P _{13} P _{14}}=\frac{\left(4 P _{13}\right)\left(4 P _{14}\right)}{ P _{13} P _{14}}=16$
