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recurrence relation problem1 Attachment(s) Hi, I would like to get help with the following recurrence relation problem. Thanks in advance. |

$P(k) = P(1)\dfrac{4^{k-1}}{k!}$ |

Thanks, but how is that explicit if the answer contains p(1)? |

I think the problem requires a boundary condition to be stated (P(1) or P(0)). |

P(1) can have any value. The formula I gave must also hold for k = 0, so it can be written as $P(k) = P(0)\dfrac{4^k}{k!}$, where P(0) can have any value. |

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