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Question: If α and β are roots of the equation \(x^{2} - ax + b = 0\) and \(V_{n} = \alpha^{n} + \beta^{n}\), ...

If α and β are roots of the equation x2−ax+b=0x^{2} - ax + b = 0 and Vn=αn+βnV_{n} = \alpha^{n} + \beta^{n}, then

A

Vn+1=aVn−bVn−1V_{n + 1} = aV_{n} - bV_{n - 1}

B

Vn+1=bVn−aVn−1V_{n + 1} = bV_{n} - aV_{n - 1}

C

Vn+1=aVn+bVn−1V_{n + 1} = aV_{n} + bV_{n - 1}

D

Vn+1=bVn+aVn−1V_{n + 1} = bV_{n} + aV_{n - 1}

Answer

Vn+1=aVn−bVn−1V_{n + 1} = aV_{n} - bV_{n - 1}

Explanation

Solution

Since α and β are roots of equation, x2−ax+b=0x^{2} - ax + b = 0, therefore α+β=a\alpha + \beta = a, αβ=b\alpha\beta = b

Now, Vn+1=αn+1+βn+1=(α+β)(αn+βn)−αβ(αn−1+βn−1)V_{n + 1} = \alpha^{n + 1} + \beta^{n + 1} = (\alpha + \beta)(\alpha^{n} + \beta^{n}) - \alpha\beta(\alpha^{n - 1} + \beta^{n - 1})

⇒ Vn+1=a.Vn−b.Vn−1V_{n + 1} = a.V_{n} - b.V_{n - 1}