A Binomial Coefficient Sum: <munderover> &#x2211;<!-- ∑ --> <mrow class="MJX-TeXAtom-O

Daphne Haney

Daphne Haney

Answered question

2022-05-15

A Binomial Coefficient Sum: m = 0 n ( 1 ) n m ( n m ) ( m 1 l )

Answer & Explanation

Kylan Simon

Kylan Simon

Beginner2022-05-16Added 17 answers

m = 0 n ( 1 ) n m ( n m ) ( m 1 l ) = ( 1 ) l + n + l + 1 m n ( 1 ) n m ( n m ) ( m 1 l )
So we need to compute this last sum. It is clearly zero if l n, so we assume lIt is equal to f(1) where f ( x ) = l + 1 m n ( 1 ) n m ( n m ) ( m 1 l ) x m 1 l . We have that
f ( x ) = 1 l ! d l d x l ( l + 1 m n ( 1 ) n m ( n m ) x m 1 ) = 1 l ! d l d x l ( ( 1 ) n + 1 x + 0 m n ( 1 ) n + 1 ( n m ) ( x ) m 1 ) = 1 l ! d l d x l ( ( 1 ) n + 1 x + ( x 1 ) n x ) = ( 1 ) n + 1 + l x l + 1 + 1 l ! k = 0 l ( l k ) n ( n 1 ) ( n k + 1 ) ( x 1 ) n k ( 1 ) l k ( l k ) ! x 1 + l k
(this last transformation thanks to Leibniz) and since n>l, f ( 1 ) = ( 1 ) l + n + 1 .
In the end, your sum is equal to ( 1 ) l + n if l n, 0 otherwise.

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