Our proposition is: 3 n > 1+2n for n =2, 3, 4, 5,... or for all n 2 Z +, n > 2. Our proposition is: 11 n 1 is divisible by 10 for all n 2 Z +

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Beryl Perkins
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1 Chapter 9 MATHEMATICAL INDUCTION EXERCISE 9A The nth term of the sequence, 7,,, 9,... is n for n Z. a () () () 7 () 9 n > n for n,,,,... or for all n Z, n > b n is divisible by 0 for all n Z c n is divisible by for all n Z d ( ) ( )( ) ( )( )( ) ( )( )( )( ) ³ ( )( )( ) :::: n n for all n Z a :::: n n(n ) for all n Z 680 nth term np ) i n(n ) for all n Z i b!!! ()!!!8!!!! where each number result is less than a factorial number c!!! 9!!!!! :::: n n! (n )! for all n Z np ) i i! (n )! for all n Z i!!!!! 6 6!!!!! 6!!!!!! !!

2 d For n Mathematics HL (rd edn), Chapter 9 MATHEMATICAL INDUCTION 67!!!! :::: n (n )! (n )! (n )! np i (n )! ) for all n Z (i )! (n )! i for all n Z 0, 6,, 8 is arithmetic with u 0, d 6 u n u (n )d 06(n ) 6n 8 8 :::: (n )(n ) n 6n for all n Z f,, 8, are arithmetic with u, d ) u n (n ) n g np ) (i )(i ) n for all n Z 6n i For n For n 7 6 T T T 7 The maximum number of triangles for n points within the original triangle is given by T n n for all n Z. a n n C C 6 b When n, C 0 n, C n, C n, C 8 n, C 6 So, from the cases n,,,,, our conjecture is: The number of regions for n points placed around a circle is given by C n n for all n Z. c n By the conjecture we expect 6 regions, but there are only. So, we no longer believe the conjecture.

3 68 Mathematics HL (rd edn), Chapter 9 MATHEMATICAL INDUCTION EXERCISE 9B. a If n 0, n 0 which is divisible by. n () n n n n n n :::: n n n n n n n n n :::: n n n n n n n n n :::: n n n n n where the contents of the brackets is an integer. ) n is divisible by. b P n is: n is divisible by for all integers n > 0. () If n 0, 0 ) P 0 is true. () If P k is true, then k A where A is an integer, and A >. Now k k (A ) 6A 6A (A ) where A is an integer as A Z Thus k is divisible by if k is divisible by. Since P 0 is true, and P k is true whenever P k is true, then P n is true for all integers n > 0 fprinciple of mathematical inductiong a If n 0, 6 n which is divisible by. 6 n ( ) n n n n n n n n n :::: n n n n n n n n n n :::: n n n n n n n n n :::: n where the contents of the brackets is an integer. ) 6 n is divisible by. b P n is: 6 n is divisible by for all integers n > 0 () If n 0, which is divisible by ) P 0 is true () If P k is true, then 6 k A where A N Now 6 k 6 6 k 6(A ) 0A 6 0A (6A ) where 6A is an integer as A N Thus, 6 k is divisible by if 6 k is divisible by. Since P 0 is true, and P k is true whenever P k is true, then P n is true for all integers n > 0 fprinciple of mathematical inductiong

4 Mathematics HL (rd edn), Chapter 9 MATHEMATICAL INDUCTION 69 a P n is: n n is divisible by for all positive integers n () If n, () which is divisible by () If P k is true, then k k A where A Z Now (k ) (k ) k k k k (k k)k k A k k (A k k ) where A k k is an integer as A and k are integers Thus (k ) (k ) is divisible by if k k is divisible by. Since P is true, and P k is true whenever P k is true, then P n is true for all positive integers n fprinciple of mathematical inductiong b P n is: n(n ) is divisible by 6 for all integers n Z () If n, ( ) 66 which is divisible by 6 ) P is true () If P k is true, then k(k )6A where A is an integer Now (k )[(k ) ](k )(k k ) (k )(k k 6) k k 6k k k 6 k k [k k 6] k(k )k(k )6 6A 6k(k ) We notice that k(k ) is the product of consecutive integers, one of which must be even ) k(k )B where B Z ) (k )[(k ) ]6A 6(B) 6(A B) where A B Z Thus (k )[(k ) ] is divisible by 6 if k(k ) is divisible by 6. Since P is true, and P k is true whenever P k is true, then P n is true for all integers n Z fprinciple of mathematical inductiong c P n is: 7 n n n is divisible by for all n Z () If n, 7 0 which is divisible by ) P is true () If P k is true, then 7 k k k A where A Z Now 7 k k k 7(7 k ) ( k ) ( k ) 7[A k k ] ( k ) ( k ) 8A 7( k ) 7( k ) ( k ) ( k ) 8A ( k ) ( k ) 8A k k (7A k k ) where k >, k Z an integer fas k and k are integersg Thus 7 k k k is divisible by if 7 k k k is divisible by. Since P is true, and P k is true whenever P k is true, then P n is true for all n Z fprinciple of mathematical inductiong

5 70 Mathematics HL (rd edn), Chapter 9 MATHEMATICAL INDUCTION EXERCISE 9B. n(n ) a P n is: :::: n for all n Z () If n, LHS and RHS (), ) P is true k(k ) () If P k is true then :::: k Thus :::: k (k ) k(k ) k k(k ) (k ) fto equalise denominatorsg (k )(k ) (k ) fcommon factor of g (k )([k ]) Since P is true, and P k is true whenever P k is true, then P n is true for all n Z fprinciple of mathematical inductiong b P n is: n(n )(n ) :::: n(n ) for all n Z () If n, LHS and RHS ()(), ) P is true () If P k is true then k(k )(k ) :::: k(k ) ) :::: k(k )(k )(k ) k(k )(k ) (k )(k ) k(k )(k ) (k )(k ) fto equalise denominatorsg (k )(k )(k ) fcommon factor of (k )(k )g [k ]([k ] )([k ]) Since P is true, and P k is true whenever P k is true, then P n is true for all n Z fprinciple of mathematical inductiong c P n is: :::: n(n ) Proof: for all n Z (By the principle of mathematical induction) () If n, LHS, RHS () If P k is true, then :::: k(k ) ( ) k(k )(k ) n(n )(n ), ) P is true

6 Mathematics HL (rd edn), Chapter 9 MATHEMATICAL INDUCTION 7 Now :::: k(k )(k )(k ) k(k )(k ) (k )(k ) k(k )(k ) 6(k )(k ) fto equalise denominatorsg (k )[k(k ) 6(k )] fcommon factorg (k )[k 9k 0] (k )(k )(k ) (k )([k ] )([k ]) Since P is true, and P k is true whenever P k is true, then P n is true for all n Z fprinciple of mathematical inductiong d P n is: :::: n n (n ) for all n Z () If n, LHS, RHS () ) P is true () If P k is true, then :::: k k (k ) Now :::: k (k ) k (k ) k (k ) (k ) (k ) fequalising denominatorsg (k ) [k (k )] fcommon factorg (k ) (k k ) (k ) (k ) Since P is true, and P k is true whenever P k is true, then P n is true for all n Z fprinciple of mathematical inductiong a The sum of the first n odd numbers 7:::: n is the sum of the first n terms of an arithmetic series. u, d. ) u n u (n )d (n ) n. Thus S n n (u (n )d) n ( (n )) n ( n ) n (n) n So, the sum of the first n odd numbers is n.

Math 55: Discrete Mathematics

Math 55: Discrete Mathematics UC Berkeley, Fall 2011 Homework # 5, due Wednesday, February 22 5.1.4 Let P (n) be the statement that 1 3 + 2 3 + + n 3 = (n(n + 1)/2) 2 for the positive integer n. a) What