(B^(k+1))=((2^k)−1)(3B−2I)−((2^k)−2)B

asked by guest
on Jan 22, 2025 at 11:06 am



You asked:

Investigate the equation: \({B}^{k + 1} = \left({2}^{k} - 1\right) \left(3 B - 2 I\right) - \left({2}^{k} - 2\right) \cdot B\).

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