Suppose a, b are two real numbers and f(n)=αn+βn. LetΔ=31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)If Δ=k(α−1)2(β−1)2(α−β)2, then k is equal to

Δ=1+1+11+α+β1+α2+β21+α+β1+α2+β21+α3+β31+α2+β21+α3+β31+α4+β4=1111αβ1α2β21111αβ1α2β2=Δ12where Δ1=1111αβ1α2β2Applying C3→C3−C2 and C2→C2−C1 we getΔ1=1001α−1β−α1α2−1β2−α2=(α−1)(β−α)11α+1β+α=(α−1)(β−1)(β−α)Thus Δ=(α−1)2(β−1)2(α−β)2∴ k=1

Suppose a, b are two real numbers and f(n)=αn+βn. LetΔ=31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)If Δ=k(α−1)2(β−1)2(α−β)2, then k is equal to

Δ=1+1+11+α+β1+α2+β21+α+β1+α2+β21+α3+β31+α2+β21+α3+β31+α4+β4=1111αβ1α2β21111αβ1α2β2=Δ12where Δ1=1111αβ1α2β2Applying C3→C3−C2 and C2→C2−C1 we getΔ1=1001α−1β−α1α2−1β2−α2=(α−1)(β−α)11α+1β+α=(α−1)(β−1)(β−α)Thus Δ=(α−1)2(β−1)2(α−β)2∴ k=1

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Suppose $a, b$ are two real numbers and $f (n) = α^{n} + β^{n}$ . Let
$Δ = |\begin{array}{ccc} 3 & 1 + f (1) & 1 + f (2) \\ 1 + f (1) & 1 + f (2) & 1 + f (3) \\ 1 + f (2) & 1 + f (3) & 1 + f (4) \end{array}|$
If $Δ = k (α - 1)^{2} (β - 1)^{2} (α - β)^{2},$ then $k$ is equal to

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$α^{2} β^{2}$

$4 α β$

answer is A.

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Detailed Solution

$\begin{array}{l} Δ = |\begin{array}{ccc} 1 + 1 + 1 & 1 + α + β & 1 + α^{2} + β^{2} \\ 1 + α + β & 1 + α^{2} + β^{2} & 1 + α^{3} + β^{3} \\ 1 + α^{2} + β^{2} & 1 + α^{3} + β^{3} & 1 + α^{4} + β^{4} \end{array}| \\ = |\begin{array}{ccc} 1 & 1 & 1 \\ 1 & α & β \\ 1 & α^{2} & β^{2} \end{array}| |\begin{array}{ccc} 1 & 1 & 1 \\ 1 & α & β \\ 1 & α^{2} & β^{2} \end{array}| = Δ_{1}^{2} \end{array}$

where $Δ_{1} = |\begin{array}{ccc} 1 & 1 & 1 \\ 1 & α & β \\ 1 & α^{2} & β^{2} \end{array}|$

Applying $C_{3} \to C_{3} - C_{2}$ and $C_{2} \to C_{2} - C_{1}$ we get

$\begin{aligned} Δ_{1} = |\begin{array}{ccc} 1 & 0 & 0 \\ 1 & α - 1 & β - α \\ 1 & α^{2} - 1 & β^{2} - α^{2} \end{array}| \\ = (α - 1) (β - α) |\begin{array}{cc} 1 & 1 \\ α + 1 & β + α \end{array}| \end{aligned}$