chapter
Comment.
Pages 1

If we replace m with n−m− 1 in the formula (9.46), then we get

( n + m− k

n− k )(

n−m− 1 + k k

) = (

2n− 1 n

) (m ≤ n− 1).

Now, if we change the summation variable, replacing k with n − k, i.e., changing the direction of summation, then we get

( m + k

k

)( 2n−m− 1− k

n− k )

= (

2n− 1 n

) (m ≤ n− 1). (9.51)

But this is the natural continuation of (9.46), so if we add (9.46) and (9.51), we get

( 2n−m− 1− k

n− k )(

m + k k

) = (

2n n

) ,

which happens to be valid for all complex values of m. This is just a simple consequence of the Chu-Vandermonde equation

(8.3). To see this we only need to change the binomial coefficients by (2.11):

( 2n−m− 1− k

n− k )(

m + k k

)

= (−1)n n∑

( n− k − 2n + m + 1 + k − 1

n− k )(

k −m− k − 1 k

)

= (−1)n n∑

( m− n n− k

)(−m− 1 k

)

= (−1)n (−n− 1

n

)

= (

2n n

) .