The base case is obvious assume true for n 1 disks

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inductive, and a minor variation on the first argument. The base case is obvious. Assume true for n - 1 disks. Now we have to prove that for n disks. At some point we have to move the largest disk. At that point all the n - 1 disks have to be on tower C. By the induction hypothesis, that cannot be done in less than 2 n - 1 - 1 moves. Then we move the largest disk to tower B. And then again move the rest of the n - 1 disks back to tower B in a minimum of 2 n - 1 - 1 moves. 5 2(a) lim t →- 2 f ( t ) = 0 2(b) lim t →- 1 f ( t ) = - 1 2(c) lim t 0 f ( t ) is not defined. 4(a) False 4(b) False 4(c) True 4(d) True 4(e) True 17(a) The table of values can be the following. θ sin θ sin θ/θ 1 0.841471 0.841471 0.1 0.099834 0.99834 0.01 0.00999983 0.999983 0.001 0.001 1 4
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