Alternate methods to solve the following summation without using the beta function [closed]

Question

$$ S = \sum_{k=0}^{n} \frac{(-1)^k}{k^3 + 9k^2 + 26k + 24} \binom{n}{k} $$ im aware that the denominator can be factored as (k+2)(k+3)(k+4). but after using partial fractions, im stuck as to how to evaluate it further.

Answer 1

Hint: Denote

$$Z(x)= \sum_{k=0}^{n} \frac{(-1)^k}{(k+2)(k+3)(k+4)} \binom{n}{k}\cdot \color{red}{x^{k+4}}$$

then $$Z'''(x)= \sum_{k=0}^{n} (-1)^k\binom{n}{k}\cdot \color{red}{x^{k+1}} = x(1-x)^n$$ then, it suffices to compute the function $Z(x)$ from its third derivative $Z'''(x)$.

Answer 2

$$ S=\sum_{k=0}^n \frac{(-)^k}{(k+2)(k+3)(k+4)}\binom{n}{k} $$ $$ =\sum_{k=0}^n \frac{(-)^k}{2(k+4)}\binom{n}{k} -\sum_{k=0}^n \frac{(-)^k}{k+3}\binom{n}{k} +\sum_{k=0}^n \frac{(-)^k}{2(k+2)}\binom{n}{k} $$ Then we have in general $$ \sum_{k=0}^N\binom{N}{k} (-1)^k\frac{1}{x+K} = \frac{N!}{x(x+1)\cdots (x+N)} $$ see https://doi.org/10.37236/1265 . Cancelling some factorials $$ S=\frac{3n!}{(4+n)!}-\frac{2n!}{(3+n)!}+\frac{n!}{2(2+n)!} = \frac{1}{2(3+n)(4+n)}. $$

Answer 3

In case you want to telescope, if we let $F(n,k)$ be your summand, we have $F(n,k)=G(n,k+1)-G(n,k)$ for $$G(n,k):=(-1)^{k+1}{n \choose k}\frac{3+k(n+4)}{(k + 3)(k + 2)(k + 1)(n + 4)(n + 3)}.$$ So your sum is equal to $$G(n,n+1)-G(n,0)=\frac{1}{2(n+3)(n+4)}.$$ The opportunity to telescope can be discovered by performing Gosper's algorithm. It can be done by hand, but there's also the chance to use Sage to do it for you. This is an example code.

var('n k')
f=(-1)^k*binomial(n,k)/(k^3+9*k^2+26*k+24)
f.gosper_sum(k)

This method can provide quick proofs of otherwise complicated identities.