Complex Numbers And Quadratic Equations question 187
Question: If $ \alpha ,\beta $ are the roots of $ x^{2}-ax+b=0 $ and if $ {{\alpha }^{n}}+{{\beta }^{n}}=V_{n} $ , then [RPET 1995; Karnataka CET 2000; Pb. CET 2002]
Options:
A) $ {V_{n+1}}=aV_{n}+b{V_{n-1}} $
B) $ {V_{n+1}}=aV_{n}+a{V_{n-1}} $
C) $ {V_{n+1}}=aV_{n}-b{V_{n-1}} $
D) $ {V_{n+1}}=a{V_{n-1}}-bV_{n} $
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Answer:
Correct Answer: C
Solution:
Multiplying $ x^{2}-ax+b=0 $ by $ {x^{n-1}} $ $ {x^{n+1}}-ax^{n}+b{x^{n-1}}=0 $ …..(i) $ \alpha ,\beta $ are roots of $ x^{2}-ax+b=0 $ , therefore they will satisfy (i) also $ {{\alpha }^{n+1}}-a{{\alpha }^{n}}+b{{\alpha }^{n-1}}=0 $ …..(ii) and $ {{\beta }^{n+1}}-a{{\beta }^{n}}+b{{\beta }^{n-1}}=0 $ …..(iii) Adding (ii) and (iii) $ ({{\alpha }^{n+1}}+{{\beta }^{n+1}})-a({{\alpha }^{n}}+{{\beta }^{n}})+b({{\alpha }^{n-1}}+{{\beta }^{n-1}})=0 $ or $ {V_{n+1}}-aV_{n}+b{V_{n-1}}=0 $ or $ {V_{n+1}}=aV_{n}-b{V_{n-1}}=0 $ (Given $ {{\alpha }^{n}}+{{\beta }^{n}}=V_{n} $ ) Trick: Put $ n=0 $ , $ 1,2 $ $ V_0={{\alpha }^{0}}+{{\beta }^{0}}=2 $ , $ V_1=\alpha +\beta =a $ , $ {{\alpha }^{2}}+{{\beta }^{2}}=V_2=a^{2}-2b $ Now the option C Þ $ V_2=aV_1-bV_0=a^{2}-2b $
BETA
