Section 12.5 AN95 7. (I) = = − n 1: 2 1 1 1 and − = 2 1 1 1 (II) If 1 2 2 2 2 1, k k 2 1 + + + + = − − then ( ) + + + + + = + + + + + ( ) − + − − 1 2 2 2 2 1 2 2 2 2 k k k k 2 1 1 1 2 1 = − + = ⋅ − = − + 2 1 2 2 2 1 2 1. k k k k 1 9. (I) = = − n 1: 4 1 1 1 and ( ) − = ⋅ = 1 3 4 1 1 3 3 1 1 (II) If ( ) + + + + = − − 1 4 4 4 1 3 4 1 , k k 2 1 then ( ) + + + + + = + + + + + ( ) − + − − 1 4 4 4 4 1 4 4 4 4 k k k k 2 1 1 1 2 1 ( ) [ ] [ ] ( ) = − + = − + ⋅ = ⋅ − = − + 1 3 4 1 4 1 3 4 1 3 4 1 3 4 4 1 1 3 4 1 . k k k k k k 1 11. (I) = ⋅ = n 1: 1 1 2 1 2 and + = 1 1 1 1 2 (II) If ( ) ⋅ + ⋅ + ⋅ + + + = + k k k k 1 1 2 1 2 3 1 3 4 1 1 1 , then ( ) ( ) ( ) [ ] ⋅ + ⋅ + ⋅ + + + + + + + k k k k 1 1 2 1 2 3 1 3 4 1 1 1 1 1 1 ( ) ( )( ) ( )( ) ( ) ( )( ) = ⋅ + ⋅ + ⋅ + + + ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ + + + = + + + + = + + + + k k k k k k k k k k k k 1 1 2 1 2 3 1 3 4 1 1 1 1 2 1 1 1 2 2 1 1 2 ( )( ) ( ) ( )( ) ( ) = + + + + = + + + = + + = + + + k k k k k k k k k k k 2 1 1 2 1 1 2 1 2 1 1 1 . 2 2 13. (I) = = n 1: 1 1 2 and ⋅ ⋅ ⋅ = 1 6 1 2 3 1 (II) If ( )( ) + + + + = + + k k k k 1 2 3 1 6 1 2 1 , 2 2 2 2 then k k 1 2 3 1 2 2 2 2 2 ( ) + + + + + + ( ) ( ) ( ) ( )( ) ( ) =++++++= + +++= + ++ k k k k k k k k k 1 2 3 1 1 6 1 2 1 1 1 6 2 9 13 6 2 2 2 2 2 2 3 2 ( )( )( ) ( ) ( ) [ ] ( ) [ ] =++ +=+ ++ ++ k k k k k k 1 6 1 2 2 3 1 6 1 1 1 2 1 1 . 15. (I) = − = n 1: 5 1 4 and 1 2 1 9 1 1 2 8 4 ( ) ⋅ ⋅ − = ⋅ = (II) If ( ) ( ) + + + + − = − k k k 4 3 2 5 1 2 9 , then ( ) ( ) [ ] + + + + − + − + k k 4 3 2 5 5 1 ( ) ( ) ( ) [ ] ( ) =++++− +−= −+−= − +− =−+ + k k k k k k k k k k 4 3 2 5 4 1 2 9 4 1 2 9 8 2 1 2 7 8 2 2 ( )( ) ( ) ( ) [ ] = + − = + − + k k k k 1 2 1 8 1 2 1 9 1 . 17. (I) ( ) = ⋅ + = n 1:1 1 1 2and ⋅ ⋅ ⋅ = 1 3 1 2 3 2 (II) If ( ) ( )( ) ⋅+⋅+⋅++ += + + k k k k k 1 2 2 3 3 4 1 1 3 1 2 , then ( ) ⋅ + ⋅ + ⋅ + + + k k 1 2 2 3 3 4 1 k k k k k k 1 1 1 1 2 2 3 3 4 1 1 2 ( ) ( ) [ ] ( ) [ ] ( )( ) + + ++=⋅+⋅+⋅++ + + + + ( )( ) ( )( ) ( )( )( ) ( ) ( ) [ ] ( ) [ ] = + + + ⋅ ++=++ +=+ ++ ++ k k k k k k k k k k k 1 3 1 2 1 3 3 1 2 1 3 1 2 3 1 3 1 1 1 1 2 . 19. (I) n 1: 1 1 2, 2 = + = which is divisible by 2. (II) If + k k 2 is divisible by 2, then ( ) ( ) ( ) + ++=++++= +++ k k k k k k k k 1 1 2 1 1 2 2. 2 2 2 Since + k k 2 is divisible by 2 and + k2 2 is divisible by ( ) ( ) + + + k k 2, 1 1 2 is divisible by 2. 21. (I) = − + = n 1: 1 1 2 2, 2 which is divisible by 2. (II) If − + k k 2 2 is divisible by 2, then ( ) ( ) ( ) + −++= + +−−+= −++ k k k k k k k k 1 1 2 2 1 1 2 2 2 . 2 2 2 Since − + k k 2 2 is divisible by 2 and 2k is divisible by 2, ( ) ( ) + − + + k k 1 1 2 2 is divisible by 2. 23. (I) = n 1: If > x 1, then = > x x 1. 1 (II) Assume, for an arbitrary natural number k, that if > x 1 then > x 1. k Multiply both sides of the inequality > x 1 k by x. If > x 1, then > > + x x 1. k 1 25. (I) = − n a b 1: is a factor of − = − a b a b. 1 1 (II) If −a b is a factor of − a b , k k then ( ) ( ) − = − + − + + a b a a b b a b . k k k k k 1 1 Since −a b is a factor of − a b k k and −a b is a factor of −a b, then −a b is a factor of − + + a b . k k 1 1 27. (a) ( ) = + = + ≥ + ⋅ n a a a 1: 1 1 1 1 1 (b) Assume that there is an integer k for which the inequality holds. So ( ) + ≥ + a ka 1 1 . k We need to show that ( ) ( ) + ≥ + + + a k a 1 1 1 . k 1 ( ) ( ) ( ) ( )( ) ( ) ( ) + =+ +≥+ +=+ ++=++ + ≥++ + a a a ka a ka a ka k a ka k a 1 1 1 1 1 1 1 1 1 1 . k k 1 2 2 29. If + + + + = + + k k k 2 4 6 2 2, 2 then ( ) ( ) ++++ + +=++++ + + k k k k 2 4 6 2 2 1 2 4 6 2 2 2 ( ) ( ) ( ) ( ) =++++=++= +++++=+ +++ k k k k k k k k k k 2 2 2 3 4 2 1 1 2 1 1 2. 2 2 2 2 But ⋅ = 2 1 2 and + + = 1 1 2 4. 2 The fact is that + + + + = + n n n 2 4 6 2 , 2 not + + n n 2 2 (Problem 1).
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