Evaluate sum_{k=0}^{n} ((2k),(k)) ((2n-2k),(n-k))

Braylon Lester 2022-07-15 Answered
Discrete math evaluate
Evaluate k = 0 n ( 2 k k ) ( 2 n 2 k n k )
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Answers (2)

Tolamaes04
Answered 2022-07-16 Author has 12 answers
Step 1
Cauchy product
Note, the binomial expression in OPs question is a Cauchy product which we get as the coefficient of the multiplication of two power series. In case both series are equal, say A ( x ) = n = 0 a k x k we obtain
(1) A 2 ( x ) = ( n 0 a n x n ) 2 = n 0 ( k = 0 n a k a n k ) x n
Applying (1) and using the hint stated in OPs question with a n = ( 2 n n ) we get
n 0 ( k = 0 n ( 2 k k ) ( 2 n 2 k n k ) ) x n = ( n 0 ( 2 n n ) x n ) 2 = ( 1 1 4 x ) 2 = 1 1 4 x = n 0 4 n x n
Step 2
We conclude by comparing coefficients
k = 0 n ( 2 k k ) ( 2 n 2 k n k ) = 4 n n 0

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Dean Summers
Answered 2022-07-17 Author has 4 answers
Step 1
Coefficient of - operator
Use the coefficient of operator [ x n ] to denote the coefficient of x n of a series A ( x ) = k 0 a k x k
Step 2
We obtain for n 0
k = 0 n ( 2 k k ) ( 2 n 2 k n k ) = k = 0 n [ x k ] 1 1 4 x [ y n k ] 1 1 4 y (2) = [ y n ] 1 1 4 y k = 0 n y k [ x k ] 1 1 4 x (3) = [ y n ] 1 1 4 y 1 1 4 y = [ y n ] 1 1 4 y = [ y n ] n 0 4 n y n = 4 n

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