If f(n)=αn+βn and 3 1+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4) =k(1−α)2(1−β)2(α−β)2, then k is equal to

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If $f (n) = α^{n} + β^{n}$ and $|\begin{array}{r} 3 & 1 + f (1) & 1 + f (2) \\ 1 + f (1) & 1 + f (2) & 1 + f (3) \\ 1 + f (2) & 1 + f (3) & 1 + f (4) \end{array}| = k {(1 - α)}^{2} {(1 - β)}^{2} {(α - β)}^{2}$ , then k is equal to

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Detailed Solution

$Δ = | \begin{matrix} 3 & a + α + β & 1 + α^{2} + β^{2} \\ 1 + α + β & 1 + α^{2} + β^{2} & 1 + α^{2} + β^{2} \\ 1 + α^{2} + β^{2} & 1 + α^{2} + β^{2} & 1 + α^{4} + β^{4} \end{matrix} | | \begin{array}{r} 1 & 1 & 1 \\ 1 & α & β \\ 1 & α^{2} & β^{2} \end{array} | \times | \begin{matrix} 1 & 1 & 1 \\ 1 & α & β \\ 1 & α^{2} & β^{2} \end{matrix} |$

Applying $C_{2} \to C_{2} - C_{1} {and C}_{3} \to C_{2} - C_{1}$

$then | \begin{array}{r} 1 & 0 & 0 \\ 1 & α - 1 & β - 1 \\ 1 & α^{2} - 1 & β^{2} - 1 \end{array} | = {(α - 1)}^{2} {(β - 1)}^{2} {(β - α)}^{2} {(1 - α)}^{2} {(1 - β)}^{2} {(α - β)}^{2} H e n c e, k = 1$