If α ,   β ≠ 0 , f n = α n + β n and 3 1 + f 1 1 + f 2 1 + f 1 1 + f 2 1 + f…

If α, β≠0, fn=αn+βn and 31+f11+f21+f11+f21+f31+f21+f31+f4=K1-α21-β2α-β2, then K is equal to 
  1. 1
  2. -1
  3. $\alpha \beta$
  4. $\frac{1}{\alpha \beta}$

Solution

$f_{1} = \alpha + \beta, f_{2} = \alpha^{2} + \beta^{2}, f_{3} = \alpha^{3} + \beta^{3}, f_{4} = \alpha^{4} + \beta^{4}$ So, the given determinant can be written as $\begin{aligned} \begin{vmatrix} 1 + 1 + 1 & 1 + \alpha + \beta & 1 + \alpha^{2} + \beta^{2} \\ 1 + \alpha + \beta & 1 + \alpha^{2} + \beta^{2} & 1 + \alpha^{3} + \beta^{3} \\ 1 + \alpha^{2} + \beta^{2} & 1 + \alpha^{3} + \beta^{3} & 1 + \alpha^{4} + \beta^{4} \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & \alpha^{2} & \beta^{2} \end{vmatrix} \begin{vmatrix} 1 & 1 & 1 \\ 1 & \alpha & \alpha^{2} \\ 1 & \beta & \beta^{2} \end{vmatrix} \end{aligned}$ $\begin{aligned}= \begin{vmatrix} 1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & \alpha^{2} & \beta^{2} \end{vmatrix}^{2} = \begin{bmatrix} \alpha \beta^{2} - \alpha^{2} \beta - \beta^{2} + \beta + \alpha^{2} - \alpha \end{bmatrix}^{2} \end{aligned}$ $\begin{aligned}= \begin{bmatrix} \alpha \beta - \beta - \alpha + 1 \end{bmatrix}^{2} = \begin{pmatrix} \beta - \alpha \end{pmatrix}^{2} \begin{pmatrix} \alpha - 1 \end{pmatrix}^{2} \begin{pmatrix} \beta - 1 \end{pmatrix}^{2} \end{aligned}$ Hence, $K = 1$.

Asked in: JEE Main 2014 (06 Apr)

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