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Potwe 13 Fc Np 10 s

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  WWW.CEMC.UWATERLOO.CA | The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the WeekProblem E and SolutionA Curious Pattern Problem Let  S  ( n ) be the sum of the first  n  terms of the series1 − 2 + 3 − 4 + 5 − 6 + 7 − 8 + 9 − 10 + 11 − 12 +  ... Show that  S  ( a ) +  S  ( b ) +  S  ( a  +  b ) = 1 when both  a  and  b  are odd. Solution a  and  b  are both odd.Let  a  = 2 k − 1 and  b  = 2 m − 1, for some positive integers  k  and  m . If   k  =  m  then  a  =  b  but wewant to prove the result true for any odd positive integers,  a  and  b .We have  S  ( a ) =  S  (2 k − 1) =  k  and  S  ( b ) =  S  (2 m − 1) =  m .Also,  a  +  b  = (2 k − 1) + (2 m − 1) = 2 k  + 2 m − 2 = 2( k  +  m − 1) is even, so S  ( a  +  b ) =  S  (2( k  +  m − 1)) = − ( k  +  m − 1) = − k − m  + 1.Then  S  ( a ) +  S  ( b ) +  S  ( a  +  b ) =  k  +  m  + ( − k − m  + 1) = 1.Therefore,  S  ( a ) +  S  ( b ) +  S  ( a  +  b ) = 1 whenever  a  and  b  are odd positive integers. Extension Since we have formulas for  S  ( a ) and  S  ( b ) based on whether  a  and  b  are even or odd, we willbreak into cases based on the parity of   a  and  b  to show that none of the other possible casesmake  S  ( a ) +  S  ( b ) +  S  ( a  +  b ) = 1 true.Case 1:  a  and  b  are both evenLet  a  = 2 k  and  b  = 2 m , for some positive integers  k  and  m .We have  S  ( a ) =  S  (2 k ) = − k  and  S  ( b ) =  S  (2 m ) = − m .Also,  a  +  b  = 2 k  + 2 m  = 2( k  +  m ) is even, so  S  ( a  +  b ) =  S  (2( k  +  m )) = − ( k  +  m ) = − k − m .The equation  S  ( a ) +  S  ( b ) +  S  ( a  +  b ) = 1 becomes  − k − m  + ( − k − m ) = 1 or 2 k  + 2 m  = − 1.Since  k  and  m  are positive integers, 2 k  + 2 m  is also a positive integer so we cannot have2 k  + 2 m  = − 1. Therefore, there is no solution.Case 2:  a  is even and  b  is oddLet  a  = 2 k  and  b  = 2 m − 1, for some positive integers  k  and  m .We have  S  ( a ) =  S  (2 k ) = − k  and  S  ( b ) =  S  (2 m − 1) =  m .Also,  a  +  b  = 2 k  + (2 m − 1) = 2( k  +  m ) − 1 is odd, so  S  ( a  +  b ) =  S  (2( k  +  m ) − 1) =  k  +  m .The equation  S  ( a ) +  S  ( b ) +  S  ( a  +  b ) = 1 becomes  − k  +  m  + ( k  +  m ) = 1 or 2 m  = 1, thus m  =  12 . But  m  must an integer, so there is no solution.Case 3:  a  is odd and  b  is evenLet  a  = 2 k − 1 and  b  = 2 m , for some positive integers  k  and  m .We have  S  ( a ) =  S  (2 k − 1) =  k  and  S  ( b ) =  S  (2 m ) = − m .Also,  a  +  b  = (2 k − 1) + 2 m  = 2( k  +  m ) − 1 is odd, so  S  ( a  +  b ) =  S  (2( k  +  m ) − 1) =  k  +  m .The equation  S  ( a ) +  S  ( b ) +  S  ( a  +  b ) = 1 becomes  k − m  + ( k  +  m ) = 1 or 2 k  = 1, thus  k  =  12 .But  k  must be an integer, so there is no solution.Therefore,  S  ( a ) +  S  ( b ) +  S  ( a  +  b ) = 1 is only true when both  a  and  b  are odd positive integers.  WWW.CEMC.UWATERLOO.CA | The CENTRE for EDUCATION in MATHEMATICS and COMPUTING We will now prove the first fact given in the question.We will show that for numbers of the form  n  = 2 k , we have  S  ( n ) =  S  (2 k ) = − k . Numbers of the form  n  = 2 k : S  (2 k ) = 1 − 2 + 3 − 4 + 5 +  ...  + (2 k − 1) − 2 k = (1 + 3 + 5 +  ...  + (2 k − 1)) − (2 + 4 + 6 +  ...  + 2 k ) grouping even and odd numbers= (1 + 3 + 5 +  ...  + (2 k − 1)) − 2(1 + 2 + 3 +  ...  +  k ) factor out a 2 from the even numbers= (1 + 3 + 5 +  ...  + (2 k − 1)) + (2 + 4 + 6 +  ...  + 2 k ) − (2 + 4 + 6 +  ...  + 2 k ) − 2(1 + 2 + 3 +  ...  +  k )add and subtract the even numbers from 2 to 2 k = (1 + 2 + 3 + 4 + 5 +  ...  + (2 k − 1) + 2 k ) − (2 + 4 + 6 +  ...  + 2 k ) − 2(1 + 2 + 3 +  ...  +  k )= (1 + 2 + 3 + 4 + 5 +  ...  + (2 k − 1) + 2 k ) − 2(1 + 2 + 3 +  ...  +  k ) − 2(1 + 2 + 3 +  ...  +  k )= (1 + 2 + 3 + 4 + 5 +  ...  + (2 k − 1) + 2 k ) − 4(1 + 2 + 3 +  ...  +  k )Now, using the formula for the sum of the first  n  integers, that is1 + 2 + 3 +  ...  + ( n − 1) +  n  =  n ( n  + 1)2we have S  (2 k ) = (1 + 2 + 3 + 4 + 5 +  ...  + (2 k − 1) + 2 k ) − 4(1 + 2 + 3 +  ...  +  k )= (2 k )(2 k  + 1)2  − 4  k ( k  + 1)2  =  k (2 k  + 1) − 2 k ( k  + 1)= 2 k 2 +  k − 2 k 2 − 2 k = − k The proof that for numbers of the form  n  = 2 k − 1, we have  S  ( n ) =  S  (2 k − 1) =  k  is similar.We’ll leave that for you to try on your own.
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