PS5_CS103

PS5_CS103 - = First, we will find a formula for this...

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========================================================== First, we will find a formula for this series. S(n) denotes the series n 2 1 ... 8 1 4 1 2 1 For small values of n, Table 1 below outlines the values of S(n) Table 1: Values of n and S(n) . n S(n) 1 2 1 2 4 3 4 1 2 1 ! 3 8 7 8 1 4 1 2 1 ! 4 16 15 16 1 8 1 4 1 2 1 ! 5 32 31 32 1 16 1 8 1 4 1 2 1 ! We propose that n n n n S 2 1 2 2 1 ... 8 1 4 1 2 1 ) ( " ! ! ========================================== We will show that n n n S 2 1 2 ) ( " ! for all . 1 # n PROOF by mathematical induction where denotes the assertion that n n n n S 2 1 2 2 1 ... 8 1 4 1 2 1 ) ( " ! ! n P BASE CASE : asserts that 1 P 2 1 2 1 2 ) 1 ( 1 1 ! " ! S Since the sum of the series is just 2 1 when there is only one term with n =1, is true. 1 P
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INDUCTIVE STEP: Assume : k P k k k k S 2 1 2 2 1 ... 8 1 4 1 2 1 ) ( " ! ! for an arbitrary integer k ı 1. Show : 1 k P 1 1 1 2 1 2 2 1 2 1 ... 8 1 4 1 2 1 ) 1 ( " ! ! k k k k k S PROOF OF THE INDUCTIVE STEP: ) P of (LHS 2 1 2 1 ... 8 1 4 1 2 1 ) 1 ( 1 k 1 ! k k k S ) 2 1 to 2 1 from terms the all (group 2 1 2 1 ... 8 1 4 1 2 1 ) 1 ( k 1 $ % & ' ± ) ! k k k S The terms in the brackets are just ). ( k S By the inductive hypothesis, replace S(k) with k k 2 1 2 " 1 1 1 1 2 1 2 2 2 2 1 2 1 2 ) 1 ( " ! " ! k k k k k k k S 1 1 2 1 2 ) 1 ( " ! k k k S Thus is true when is true, and therefore is true for all by the principle of mathematical induction. 1 k P k P n P 1 # n Q.E.D. =======================================
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========================================================== We will show that for all . 1 3 1 2 " ± n n 3 # n PROOF by mathematical induction where n P denotes the assertion that 1 3 1 2 " ± n n BASE CASE : asserts that: 3 P . 1 ) 3 ( 3 1 ) 3 ( 2 " ± 8 7 ± Since it is true that , is true. 8 7 ± 3 P INDUCTIVE STEP : Assume : for an arbitrary integer . k P 1 3 1 2 " ± k k 3 # k Show : 1 k P 1 ) 1 ( 3 1 ) 1 ( 2 " ± k k Proof of the Inductive Step: true) is P , hypothesis inductive (By the 1 3 1 2 k " ± k k sides) both to 2 add we if holds still inequality (The 2 1 3 2 1 2 " ± k k algebra) by ( 1 3 1 ) 2 2 ( ± k k algebra) by ( 1 3 1 ) 1 ( 2 ± k k RHS) to 1 add we if holds still inequality The ( 1 1 3 1 ) 1 ( 2 ± k k algebra) by ( 1 3 3 1 ) 1 ( 2 " ± k k algebra) by ( 1 ) 1 ( 3 1 ) 1 ( 2 " ± k k Thus is true when is true, and therefore is true for all by the principle of mathematical induction. 1 k P k P n P 3 # n Q.E.D. ==================================================
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========================================================== We call this game the “stone” game. For all , we will show that the second player can always win in a “stone” game with n stones in each of the two piles initially. 1 # n PROOF by mathematical induction where n P denotes the assertion that the second player can always win in a “stone” game with n stones in each of the two piles initially. BASE CASE : asserts that the second player can always win in a “stone” game with one stone in each of the two piles initially. When the first player moves, he will take one stone from one of the two pile. We will call this pile A. After the first player takes the only one stone from pile A, there will be zero stone left in pile A. Therefore, the second player will need to take a stone from the other pile. We call this pile B.
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PS5_CS103 - = First, we will find a formula for this...

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