Denote the sum S n := ∑ k = 0 n ( n k )...

Mylee Underwood

Mylee Underwood

Answered

2022-07-01

Denote the sum
S n := k = 0 n ( n k ) ( 1 ) n k n + k + 1

Answer & Explanation

Camron Herrera

Camron Herrera

Expert

2022-07-02Added 16 answers

We have
0 1 ( x ( 1 x ) ) n   dx = 0 1 ( x x 2 ) n   dx = 0 1 k = 0 n ( n k ) x k ( x 2 ) n k   dx = k = 0 n ( n k ) ( 1 ) n k 0 1 x 2 n k   dx = k = 0 n ( n k ) ( 1 ) n k 2 n k + 1 = k = 0 n ( n n k ) ( 1 ) k 2 n ( n k ) + 1 = k = 0 n ( n k ) ( 1 ) k n + k + 1 .
where in step ( ) we use the substitution k n k
Kristen Stokes

Kristen Stokes

Expert

2022-07-03Added 2 answers

To evaluate the sum
k = 0 n ( n k ) ( 1 ) n k n + k + 1
introduce the function
f ( z ) = n ! n + 1 + z q = 0 n 1 z q
which has the property that for 0 k n
R e s z = k f ( z ) = n ! n + 1 + k q = 0 k 1 1 k q q = k + 1 n 1 k q = n ! n + 1 + k 1 k ! ( 1 ) n k ( n k ) ! = ( n k ) ( 1 ) n k n + 1 + k .
Now with residues summing to zero and the residue at infinity being zero by inspection we get for our sum the value
R e s z = n 1 f ( z ) = n ! q = 0 n 1 n 1 q = n ! ( 1 ) n q = 0 n 1 n + 1 + q = ( 1 ) n n ! n ! ( 2 n + 1 ) ! .
This is
( 1 ) n 2 n + 1 ( 2 n n ) 1 .

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