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(1−α)2(1−β)2(α−β)2, then k is equal to
∣
∣∣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(1−α)2(1−β)2(α−β)2, then k is equal to
Answer: Option A
:
A
Δ=∣∣
∣
∣∣31+α+β1+α2+β21+α+β1+α2+β21+α3+β31+α2+β21+α3+β31+α4+β4∣∣
∣
∣∣=∣∣
∣∣1111αβ1α2β2∣∣
∣∣×∣∣
∣∣1111αβ1α2β2∣∣
∣∣
Applying C2→−C2−C3→C3−C1,
∣∣
∣∣1001α−1β−11α2−1β2−1∣∣
∣∣2=(α−1)2(β−1)2(β−α)2=(1−α)2(1−β)2(α−β)2
Hence, k=1
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:
A
Δ=∣∣
∣
∣∣31+α+β1+α2+β21+α+β1+α2+β21+α3+β31+α2+β21+α3+β31+α4+β4∣∣
∣
∣∣=∣∣
∣∣1111αβ1α2β2∣∣
∣∣×∣∣
∣∣1111αβ1α2β2∣∣
∣∣
Applying C2→−C2−C3→C3−C1,
∣∣
∣∣1001α−1β−11α2−1β2−1∣∣
∣∣2=(α−1)2(β−1)2(β−α)2=(1−α)2(1−β)2(α−β)2
Hence, k=1
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