Ononot — The recurrence an = 4a{n-1} - 3 \cdot 2^n has homogeneous so

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Recurrence Relationsmedium

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The recurrence $a_n = 4a_{n-1} - 3 \cdot 2^n$ has homogeneous solution $a_n^{(h)} = c \cdot 4^n$ . With $a_0 = 5$ , what is the general solution?