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Recurrence Relationseasy
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For the recurrence an=4an−1−4an−2a_n = 4a_{n-1} - 4a_{n-2}an​=4an−1​−4an−2​ with a0=1a_0 = 1a0​=1 and a1=2a_1 = 2a1​=2, the characteristic equation gives a repeated root r=2r = 2r=2. The general solution is an=(A+Bn)⋅2na_n = (A + Bn) \cdot 2^nan​=(A+Bn)⋅2n. Find AAA and BBB.