Explanation The given recurrence relation is, s n = s n − 1 + ( 2 n − 1 ) 2 for n ≥ 2 with the initial condition s 1 = 1 . Substitute n = 1 in s n = C ( 2 n + 1 , 3 ) and obtain s 1 as follows. s 1 = C ( 2 ( 1 ) + 1 , 3 ) = C ( 3 , 3 ) = 3 ! 3 ! ( 3 − 3 ) ! = 1 0 ! = 1 Hence the result s n = C ( 2 n + 1 , 3 ) is correct for n = 1 . Now assume the formula is correct for n = 2 , 3 , … , k . Then by assumption, s k = C ( 2 k + 1 , 3 ) Prove the formula is correct for n = k + 1 as follows. s k + 1 = s k + 1 − 1 + ( 2 ( k + 1 ) − 1 ) 2 = s k + ( 2 k + 1 ) 2 = C ( 2 k + 1 , 3 ) + ( 2 k + 1 ) 2 = ( 2 k + 1 ) ! 3 ! ( 2 k − 2 ) ! + ( 2 k + 1 ) 2 Take least common divisor and simplify further as shown below. s k + 1 = ( 2 k + 1 ) ! + ( 2 k + 1 ) 2 ( 3 ! ( 2 k − 2 ) ! ) 3 ! ( 2 k − 2 ) ! = ( ( 2 k ) ( 2 k − 1 ) ( 2 k ) ( 2 k − 1 ) ) ( ( 2 k − 2 ) ! ( 2 k − 1 ) ( 2 k ) ( 2 k + 1 ) + ( 2 k + 1 ) 2 ( 3 ! ( 2 k − 2 ) ! ) 3 ! ( 2 k − 2 ) ! ) = ( 2 k − 2 ) ! ( 2 k ) ( 2 k − 1 ) 3 ! ( 2 k ) ! ( ( 2 k ) ( 2 k − 1 ) ( 2 k + 1 ) + 6 ( 2 k + 1 ) 2 ) = ( 2 k ) !

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Author: Dossey, John A.

Publisher: Pearson,

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Chapter 9.2, Problem 8E

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To prove: The term C(2n+1,3) is a solution to the recurrence relation sn=sn−1+(2n−1)2 for n≥2 with s1=1 by mathematical induction.

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Chapter 9.2, Problem 7E

Chapter 9.2, Problem 9E

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To prove: The term C ( 2 n + 1 , 3 ) is a solution to the recurrence relation s n = s n − 1 + ( 2 n − 1 ) 2 for n ≥ 2 with s 1 = 1 by mathematical induction.

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To prove: The term C(2n+1,3) is a solution to the recurrence relation sn=sn−1+(2n−1)2 for n≥2 with s1=1 by mathematical induction.

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Chapter 9 Solutions