Proving a binomial sum identity ∑ k = 0 n ( n k ) (...

Jonathan Miles

Jonathan Miles

Answered

2022-07-01

Proving a binomial sum identity k = 0 n ( n k ) ( 1 ) k 2 k + 1 = ( 2 n ) ! ! ( 2 n + 1 ) ! ! .

Answer & Explanation

Sanaa Hinton

Sanaa Hinton

Expert

2022-07-02Added 15 answers

You could consider the integral
0 1 ( 1 x 2 ) n d x .
EnvivyEvoxys6

EnvivyEvoxys6

Expert

2022-07-03Added 7 answers

S n = k = 0 n ( 1 ) k ( n k ) 2 k + 1
We have:
S n = ( 1 ) n 2 n + 1 + k = 0 n 1 ( 1 ) k ( n k ) 2 k + 1 = ( 1 ) n 2 n + 1 + k = 0 n 1 [ n n k . ( 1 ) k ( n 1 k ) ( 2 k + 1 ) ] = ( 1 ) n 2 n + 1 + 1 2 n + 1 k = 0 n 1 [ n ( 2 n + 1 ) n k . ( 1 ) k ( n 1 k ) ( 2 k + 1 ) ] = ( 1 ) n 2 n + 1 + 1 2 n + 1 k = 0 n 1 [ 2 n 2 2 n k + 2 n k + n n k . ( 1 ) k ( n 1 k ) ( 2 k + 1 ) ] = ( 1 ) n 2 n + 1 + 1 2 n + 1 k = 0 n 1 [ 2 n 2 2 n k n k . ( 1 ) k ( n 1 k ) ( 2 k + 1 ) ] + 1 2 n + 1 k = 0 n 1 [ 2 n k + n n k . ( 1 ) k ( n 1 k ) ( 2 k + 1 ) ] = ( 1 ) n 2 n + 1 + 2 n 2 n + 1 k = 0 n 1 ( 1 ) k ( n 1 k ) ( 2 k + 1 ) + 1 2 n + 1 k = 0 n 1 n ( 1 ) k ( n 1 k ) n k = 2 n 2 n + 1 k = 0 n 1 ( 1 ) k ( n 1 k ) ( 2 k + 1 ) + 1 2 n + 1 k = 0 n 1 [ ( 1 ) k ( n k ) ] + ( 1 ) n 2 n + 1 = 2 n 2 n + 1 k = 0 n 1 ( 1 ) k ( n 1 k ) ( 2 k + 1 ) + 1 2 n + 1 k = 0 n [ ( 1 ) k ( n k ) ]
Therefore,
S n = 2 n 2 n + 1 S n 1 + 0 S n 1 = 2 n 2 2 n 1 S n 2 . . . S 1 = 2 3 S 0 and S 0 = 1
Hence,
S n = 2 n ( 2 n 2 ) . . .2 ( 2 n + 1 ) ( 2 n 1 ) . . .3 .1 = ( 2 n ) ! ! ( 2 n + 1 ) ! !

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