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Question

If f(n)=αn+βn and ∣ ∣ ∣31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)∣ ∣ ∣=k(1α)2(1β)2(αβ)2, then k is equal to
  1. 1
  2. -1
  3. αβγ
  4. αβ


Solution

The correct option is A 1
Δ=∣ ∣ ∣31+α+β1+α2+β21+α+β1+α2+β21+α3+β31+α2+β21+α3+β31+α4+β4∣ ∣ ∣   =∣ ∣1111αβ1α2β2∣ ∣×∣ ∣1111αβ1α2β2∣ ∣
Applying C2C2C3C3C1,
        ∣ ∣1001α1β11α21β21∣ ∣2    =(α1)2(β1)2(βα)2    =(1α)2(1β)2(αβ)2
Hence, k=1

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