Evaluate this simple sum!

Nice problem, but note that one might say that the following "solution" I present is also known as cheating . So read at your own risk.

Note that, if this problem is not flawed, the relationship given is true and it holds for all $n \ge 0$ . There are three unknowns: $\alpha$ , $\beta$ , and $\gamma$ , so if we have three independent equations relating $\alpha$ , $\beta$ , and $\gamma$ , we can solve for them.

For $n = 0$ , the sum is $\displaystyle\sum_{k=0}^{0} \dfrac {\left( -1 \right)^k}{k^3 + 9k^2 + 26k + 24} \dbinom {0}{k} = \dfrac {1}{24} = \dfrac {1}{\alpha \beta \gamma}.$

For $n = 1$ , the sum is $\displaystyle\sum_{k=0}^{1} \dfrac {\left( -1 \right)^k}{k^3 + 9k^2 + 26k + 24} \dbinom {1}{k} = \dfrac {1}{24} - \dfrac {1}{60} = \dfrac {1}{40} = \dfrac {1}{\alpha \left( 1 + \beta \right) \left( 1 + \gamma \right)}.$

For $n = 2$ , the sum is $\displaystyle\sum_{k=0}^{2} \dfrac {\left( -1 \right)^k}{k^3 + 9k^2 + 26k + 24} \dbinom {2}{k} = \dfrac {1}{24} - \dfrac {2}{60} + \dfrac {1}{120} = \dfrac {1}{60} = \dfrac {1}{\alpha \left( 2 + \beta \right) \left( 1 + \gamma \right)}.$

From these, we get the set of equations: $\begin{aligned} \alpha \beta \gamma &=& 24, \\ \alpha \left( 1 + \beta \right) \left( 1 + \gamma \right) &=& 40, \\ \alpha \left( 2 + \beta \right) \left( 2 + \gamma \right) &=& 60. \end{aligned}$

At this point, you realize that there are several integer solutions but $\alpha > 1$ is probably the case, so $\alpha = 2$ and $\beta + \gamma = 7$ , giving the answer $\alpha + \beta + \gamma = \boxed {9}.$

It also turns out that this is correct, using WolframAlpha: $\displaystyle\sum_{k=0}^{n} \dfrac {\left( -1 \right)^k}{k^3 + 9k^2 + 26k + 24} \dbinom {n}{k} = \dfrac {1}{2n^2 + 14n + 24}.$

Finally, a note to @Abhishek Singh - it is a good habit to name your problems for descriptively. First of all, the title says nothing about the problem itself. Second, it makes people who cannot solve the problem feel bad, since you are calling it "simple" - it may be simple for you but it's definitely not simple for everybody.

Two key elements are the partial fraction decomposition $\frac{1}{(k+2)(k+3)(k+4)}=\frac{1/2}{k+2}-\frac{1}{k+3}+\frac{1/2}{k+4}$ and the identity: $\sum_{k=0}^n\frac{(-1)^k}{k+\eta+1}\binom{n}{k}=\int_{0}^{1}x^\eta\sum_{k=0}^{n}\binom{n}{k}(-x)^k\,dx=\int_{0}^{1}x^\eta (1-x)^n = \frac{\eta!}{(n+\eta+1)\cdot\ldots\cdot(n+1)}.$ This two (pretty advanced) ingredients give that the original sum is just $\frac{1}{2(n+3)(n+4)}.$

$\sum_{k=0}^{n}\frac{(-1)^{k}}{k^{3}+9k^{2}+26k+24}(_{k}^{n})$

$=\sum_{k=0}^{n}\frac{(-1)^{k}}{(k+2)(k+3)(k+4)}(_{k}^{n})$

$=\sum_{k=0}^{n}(-1)^{k}\frac{k+1}{(n+1)(n+2)(n+3)(n+4)}(_{k+4}^{n+4})$

$=\frac{1}{(n+1)(n+2)(n+3)(n+4)}\sum_{k=4}^{n}(-1)^{k}(k-3)(_{k}^{n+4})$

$\sum_{k=0}^{n+4}(-1)^{k}(k-3)(_{k}^{n+4})$

$=\sum_{k=0}^{n+4}(-1)^{k}k(_{k}^{n+4})-3\sum_{k=0}^{n+4}(-1)^{k}(_{k}^{n+4})$

$=\sum_{k=1}^{n+4}(-1)^{k}k(_{k}^{n+4})-3(1-1)^{n+4}$

$=\frac{1}{n+4}\sum_{k=1}^{n+4}(-1)^{k}(_{k-1}^{n+3})$

$=\frac{1}{n+4}(1-1)^{n+3}=0$

Hence, $\sum_{k=4}^{n}(-1)^{k}(k-3)(_{k}^{n+4})$

$=-\sum_{k=0}^{3}(-1)^{k}(k-3)(_{k}^{n+4})$

$=3(_{0}^{n+4})-2(_{1}^{n+4})+(_{2}^{n+4})=\frac{(n+1)(n+2)}{2}$