If α,β ≠ 0 and f(n)=αn+βn and | 3&1& f(1)& 1 &f(2) [0.3em] 1 f(1)&1 &f(2)&1&f(3) [0.3em] 1&f(2)&1&f(3)&1&f(4) | =k(1-α)2(1-β)2(α-β)2 , then k is equal to

Question

If α,β ≠ 0 and f(n)=αn+βn and | 3&1& f(1)& 1 &f(2) [0.3em] 1 f(1)&1 &f(2)&1&f(3) [0.3em] 1&f(2)&1&f(3)&1&f(4) | =k(1-α)2(1-β)2(α-β)2 , then k is equal to (A) 1 (B) - 1 (C) α β (D) (1/α β)

Tardigrade Admin · Accepted Answer

Given, f(n)=αn+βn, f(1)=α+β, f(2)=α2+β2, f(3)=α3+β3, f(4)=α4+β4 Let Δ=|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)| ⇒ Δ=|3 1+α+β &1+α2+β2 1+α+β 1+α2+β2 1+α3+β3 1+α2+β2 1+α3+β3 1+α4+β4| |1⋅1+1⋅1+1⋅1&1⋅1+1⋅α+1⋅β&1⋅1+1⋅α2+1⋅β2 1⋅1+⋅α+1⋅β&1⋅1+α⋅α+β⋅β&1⋅1+α⋅α2+β⋅β2 1⋅1+1⋅α2 + 1⋅β2&1⋅1+α2⋅α+β2⋅β&1⋅1+α2⋅α2+β2⋅β2| =|1 1 1 1 α β 1 α2 β2||1 1 1 1 α β 1 α2 β2|=|1 1 1 1 α β 1 α2 β2|2 On expanding, we get Δ=(1-α)2(1-β)2(α-β)2 But given, Δ=K(1-α)2(1-β)2(α-β)2 Hence, K(1-α)2(1-β)2(α-β)2=(1-α)2(1-β)2(α-β)2 ∴ K=1

Q. If $α, β \neq = 0$ and $f (n) = α^{n} + β^{n}$ and $∣ ∣ 11 3 f (1) f (2) 111 f (1) f (2) f (3) 111 f (2) f (3) f (4) ∣ ∣$
$= k (1 - α)^{2} (1 - β)^{2} (α - β)^{2}$ , then $k$ is equal to

1

- 1

$α β$

$\frac{1}{α β}$

Q. If α,β=0 and f(n)=αn+βn and ∣∣​11​3f(1)f(2)​111​f(1)f(2)f(3)​111​f(2)f(3)f(4)​∣∣​ =k(1−α)2(1−β)2(α−β)2 , then k is equal to

1

- 1

αβ

αβ1​

Q. If $α, β \neq = 0$ and $f (n) = α^{n} + β^{n}$ and $∣ ∣ 11 3 f (1) f (2) 111 f (1) f (2) f (3) 111 f (2) f (3) f (4) ∣ ∣$
$= k (1 - α)^{2} (1 - β)^{2} (α - β)^{2}$ , then $k$ is equal to

$α β$

$\frac{1}{α β}$