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aggierabz2006zw 2022-07-03 Answered
Prove k = 0 n ( 2 k ) ! ( 2 n 2 k ) ! 2 2 n ( 2 k + 1 ) [ k ! ( n k ) ! ] 2 = [ 2 n n ! ] 2 ( 2 n + 1 ) ! using mathematical induction
Since I already know the way using combinatorial proof, I am trying to prove it by mathematical induction but it is not easy to show.
I derived that
S n = 2 n 2 n + 1 S n 1 from RHS, but how can I proceed from this? I mean, how can I organize the LHS? I cannot organize LHS properly because of different n , n = m   and   n = m + 1.
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Answers (1)

Maggie Bowman
Answered 2022-07-04 Author has 14 answers
Step 1
First of all notice that:
( 2 n + 1 ) ! = 1 2 3 4 2 n ( 2 n + 1 ) = [ 1 3 5 ( 2 n + 1 ) ] 2 n [ 1 2 n ] = ( 2 n + 1 ) ! ! 2 n n !
Step 2
Based on the first hint, you want to prove that:
S n = [ 2 n n ! ] 2 ( 2 n + 1 ) ! = 2 n n ! ( 2 n + 1 ) ! !
Step 3
What does the second hint and the rule:
S n = 2 n 2 n + 1 S n 1 have in common?
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