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Homework_Set_2_Solution

Homework_Set_2_Solution - Homework 2 Solutions 1 Apostol...

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Homework 2 Solutions October 10, 2005 1 Apostol page 45 number 8 Show n k =1 (1 + x 2 k - 1 ) = 1 - x 2 n 1 - x . Induct on n . We check that P (1) is true: Substituting k = 1 into the equation we get 1 + x = 1 - x 2 1 - x , which is true. Assume P ( n ), i.e. that n k =1 (1 + x 2 k - 1 ) = 1 - x 2 n 1 - x . Want to show that P ( n ) implies P ( n + 1). We note that n +1 k =1 (1 + x 2 k - 1 ) = ( n k =1 (1 + x 2 k - 1 )) * (1 + x 2 n ) . Use the induction hypothesis to get n +1 k =1 (1 + x 2 k - 1 ) = ( 1 - x 2 n 1 - x ) * (1 + x 2 n ) = 1 - ( x 2 n ) 2 1 - x = 1 - x 2 n +1 1 - x To evaluate n k =1 (1 + x 2 k - 1 ) at x = 1 we substitute 1 for x n k =1 (1 + x 2 k - 1 ) = n k =1 (1 + 1 2 k - 1 ) = n k =1 (1 + 1) = 2 n . 2 Apostol page 45 number 13 1
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2.1 Show that b p - a p = ( b - a )( b p - 1 + b p - 2 a + ... + a p - 1 ) = ( b - a ) p i =1 b i - 1 a p - i . By distributive laws, we have ( b - a ) p i =1 b i - 1 a p - i = b ( p i =1 b i - 1 a p - i ) - a ( p i =1 b i - 1 a p - i ) = ( p i =1 b i a p - i ) - ( p i =1 b i - 1 a p - i +1 ) . Rewriting, we get ( b - a ) p i =1 b i - 1 a p - i = b p + ( p - 1 i =1 b i a p - i ) - a p - ( p i =2 b i - 1 a p - i +1 ) . We shift the indices by letting i = j + 1 and substituting into the second sum in the above expression ( b - a ) p i =1 b i - 1 a p - i = b p + ( p - 1 i =1 b i a p - i ) - a p - ( j +1= p j +1=2 b ( j +1) - 1 a p - ( j +1)+1 ) = = b p + ( p - 1 i =1 b i a p - i ) - a p - ( j = p - 1 j =1 b j a p - j ) .
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