The problem here is that while \(P(0)\) is true, and while \(P(k) \imp P(k+1)\) for some values of \(k\text{,}\) there is at least one value of \(k\) (namely \(k = 99\)) when that implication fails. For a valid proof by induction, \(P(k) \imp P(k+1)\) must be true for all values of \(k\) greater than or equal to the base case.
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