a. Explanation Given: The set of identical dominos are 2 inches long and each domino overhangs the one beneath. Calculation: From the given Figure, the center of gravity of any stack of dominos is the average of the placed of their centers. Consider the midpoint of the top domino is x = 0 . In the Figure, the stack additional dominos down to its right. Let m ( n ) be the x -coordinate of the midpoint of the n th domino. Then the order of the stack placed directly under the center gravity of the dominos 0 through n − 1 . That is, 1 n ∑ i = 0 n − 1 m ( i ) . Therefore, the midpoint of the n th domino is m ( n ) = 1 + 1 n ∑ i = 0 n − 1 m ( i ) (1) To prove : The midpoint of the n th domino m ( n ) = ∑ k = 1 n 1 k . Base case : n = 1 . To prove that the claim is true when n = 1 . Here, m ( 1 ) = 1 + 1 1 ∑ i = 0 1 − 1 m ( i ) = 1 + 1 ( m ( 0 ) ) = 1 [ ∵ m ( 0 ) = 0 ] Therefore, the claim is when n = 1 . Induction hypothesis: n = r − 1 Assume that the claim is true when n = r − 1 . m ( r ) = ∑ k = 1 r − 1 1 k Induction step: n = r To prove that the claim is true when n = r . Substitute r for n in equation (1), m ( r ) = 1 + 1 r ∑ i = ` r − 1 m ( i ) = 1 + 1 r ∑ i = 1 r − 1 ∑ j = 1 i 1 j Here 1 appears r − 1 times in the double summation and 2 appears r − 2 times , and so on b. To determine To find: The greatest distance that the top domino for infinitely many dominoes in the stack

2nd Edition

ISBN: 9781323178522

Author: Briggs

Publisher: PEARSON

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Chapter 8.4, Problem 71E

To determine

To find: The greatest distance that the top domino can be made to overhang the bottom domino.

To determine

To find: The greatest distance that the top domino for infinitely many dominoes in the stack

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Chapter 8.4, Problem 70E

Chapter 8.4, Problem 72E

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