If α, β ≠ 0, and 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

Question

Accepted Answer

1. The determinant of the given matrix simplifies to a form that reveals K as a constant factor, specifically 1, when considering the structure of the functions involved and their relationships. This is consistent with the properties of determinants and the specific forms of f(n) in the context of the problem.

If α, β ≠ 0, and 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

Solution

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