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

Question

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

I watched a challenging high school-level problem and its solution on YouTube.

Evaluate the following sum $$\sum_{k=0}^{n}\frac{(-1)^k}{k^3+9k^2+26k+24}\binom{n}{k}.$$

I could not solve this problem but I understood the solution by Jiro Fukuda.
The solution by Jiro Fukuda is complicated. Since the result turns out to be very simple, I suspect that there might be a more natural and straightforward solution. Please show me a simple and elegant solution.

Answer 1

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

We have some nice factors,

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

We can do a partial fraction split,

$$\frac{1}{(k+2)(k+3)(k+4)}=\frac12\cdot\frac{1}{k+2} - \frac{1}{k+3} + \frac12\cdot\frac{1}{k+4}$$

Therefore,

$$S=\frac12\sum_{k=0}^{n}\frac{(-1)^k}{k+2}\binom{n}{k} - \sum_{k=0}^{n}\frac{(-1)^k}{k+3}\binom{n}{k} + \frac12\sum_{k=0}^{n}\frac{(-1)^k}{k+4}\binom{n}{k}$$

Now, consider the binomial expansion of $(1-x)^n$ and multiply by $x^{m-1}$ throughout

$$(1-x)^n \cdot x^{m-1} = \sum_{k=0}^n \binom{n}{k} (-1)^k x^k\cdot x^{m-1} $$

Integrate both sides,

$$\int_0^1(1-x)^n x^{m-1} \,dx= \sum_{k=0}^n \binom{n}{k} (-1)^k \int_0^1 x^{k+m-1}\,dx $$

$$\int_0^1(1-x)^n x^{m-1} \,dx= \sum_{k=0}^n \frac{(-1)^k}{k+m}\binom{n}{k} $$

Using the definition of Beta function to solve the integral on the left,

$$\therefore\boxed{\beta(m,n+1)=\sum_{k=0}^n \frac{(-1)^k}{k+m}\binom{n}{k}}$$

Using the symmetric property,

$$S=\frac12\beta(n+1,2) - \beta(n+1,3) + \frac12\beta(n+1,4)$$

Now simplifying the expression above, by the use of below properties,

$$\beta(n,m)=\frac{\Gamma(n)\Gamma(m)}{\Gamma(n+m+1)}, \Gamma(n+1)=n!\tag1$$

$$ \begin{aligned} &\implies \frac{1}{2} \cdot \frac{\Gamma(n+1)\Gamma(2)}{\Gamma(n+3)} - \frac{\Gamma(n+1)\Gamma(3)}{\Gamma(n+4)} + \frac{1}{2} \cdot \frac{\Gamma(n+1)\Gamma(4)}{\Gamma(n+5)} \\ &= \Gamma(n+1) \left( \frac{1}{2\Gamma(n+3)} - \frac{2}{\Gamma(n+4)} + \frac{3}{\Gamma(n+5)} \right) \end{aligned} $$

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$$ \begin{aligned} \Gamma(n+3) &= (n+2)(n+1)\Gamma(n+1) \\ \Gamma(n+4) &= (n+3)(n+2)(n+1)\Gamma(n+1) \\ \Gamma(n+5) &= (n+4)(n+3)(n+2)(n+1)\Gamma(n+1) \end{aligned} $$

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$$ {S = \frac{n^2+3n+2}{2(n+1)(n+2)(n+3)(n+4)}} $$

$$S\underset{n^2+3n+2=(n+1)(n+2)}{\implies}\frac{1}{2(n+3)(n+4)}$$

Answer 2

Hint:

From $\displaystyle(k+1)\binom{n+4}{k+4}=(k+4-3)\binom{n+4}{k+4}=(n+4)\binom{n+3}{k+3}-3\binom{n+4}{k+4}$

$\displaystyle T_2=\sum_{k=0}^n(-1)^k\binom{n+4}{k+4}=\sum_{r=4}^{n+4}(-1)^r\binom{n+4}r$ as $(-1)^k=(-1)^r$

$\displaystyle T_2=(1-1)^{n+4}-\sum_{r=0}^3\binom{n+4}r=?$

$\displaystyle T_1=\sum_{k=0}^n(-1)^k\binom{n+3}{k+3}=-\sum_{m=3}^n(-1)^m\binom{n+3}m$ as $(-1)^k=-(-1)^m$

$\displaystyle T_1=-(1-1)^{n+3}+\sum_{m=0}^2(-1)^m\binom{n+3}m=?$

Answer 3

Claude Leibovici, thank you very much for your following formula.

$$\sum_{k=0}^n (-1)^k\frac{\binom{n}{k}}{k+a}=\frac{1}{a\binom{n+a}{a}}$$ holds for any non-negative integer $n$ and for any positive integer $a$.

Proof. Let $$S_{n,a}:= \sum_{k=0}^n (-1)^k\frac{\binom{n}{k}}{k+a}.$$

I prove the above formula by induction on $a$.
(1)
Let $n$ be any non-negative integer.
Let $a=1$.

Fact 1:
$(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\dots +\binom{n}{n}x^n$ holds.
So, $0=(1-1)^n=\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\dots + (-1)^n \binom{n}{n}$ holds.
So, $1=\binom{n}{0}= \binom{n}{1}-\binom{n}{2}+\dots + (-1)^{n-1} \binom{n}{n}$ holds.

Fact 2:
$n\binom{n-1}{k}=n\frac{(n-1)!}{(n-1-k)!k!}=(k+1)\frac{n!}{(n-1-k)!(k+1)!}=(k+1)\binom{n}{k+1}$.

By Fact 2, $(n+1)\binom{n}{k}=(k+1)\binom{n+1}{k+1}$.
So, $\frac{\binom{n}{k}}{k+1}=\frac{\binom{n+1}{k+1}}{n+1}$ holds.
So, $$S_{n,1}=\sum_{k=0}^n (-1)^k\frac{\binom{n}{k}}{k+1}=\sum_{k=0}^n (-1)^k\frac{\binom{n+1}{k+1}}{n+1}=\frac{1}{n+1}\sum_{k=0}^n (-1)^k \binom{n+1}{k+1}\\=\frac{1}{n+1}\left(\binom{n+1}{1}-\binom{n+1}{2}+\binom{n+1}{3}-\dots+(-1)^n\binom{n+1}{n+1}\right).$$
By Fact 1, $S_{n,1}=\frac{1}{n+1}=\frac{1}{1\binom{n+1}{1}}$.
So, when $a=1$, the above formula holds.

(2)
Assume that $$S_{n,a}=\sum_{k=0}^n (-1)^k\frac{\binom{n}{k}}{k+a}=\frac{1}{a\binom{n+a}{a}}$$ holds for any non-negative integer $n$ and for $a$ which is greater than or equal to $1$.
$$S_{n,a}=\frac{\binom{n}{0}}{a}-\frac{\binom{n}{1}}{a+1}+\frac{\binom{n}{2}}{a+2}-\dots+(-1)^{n-1}\frac{\binom{n}{n-1}}{a+(n-1)}+(-1)^{n}\frac{\binom{n}{n}}{a+n}.$$
$$S_{n,a+1}=\frac{\binom{n}{0}}{a+1}-\frac{\binom{n}{1}}{a+2}+\frac{\binom{n}{2}}{a+3}-\dots+(-1)^{n-1}\frac{\binom{n}{n-1}}{a+n)}+(-1)^{n}\frac{\binom{n}{n}}{a+n+1}.$$ $$S_{n,a+1}-S_{n,a}=\\-\frac{\binom{n}{0}}{a}+\frac{\binom{n}{0}+\binom{n}{1}}{a+1}-\frac{\binom{n}{1}+\binom{n}{2}}{a+2}+\dots+(-1)^{n-1}\frac{\binom{n}{n-1}+\binom{n}{n}}{a+n}+(-1)^n\frac{\binom{n}{n}}{a+n+1}\\=-\frac{\binom{n+1}{0}}{a}+\frac{\binom{n+1}{1}}{a+1}-\frac{\binom{n+1}{2}}{a+2}+\dots+(-1)^{n-1}\frac{\binom{n+1}{n}}{a+n}+(-1)^n\frac{\binom{n+1}{n+1}}{a+n+1}\\=-S_{n+1,a}.$$
So, $$S_{n,a+1}=S_{n,a}-S_{n+1,a}=\frac{1}{a\binom{n+a}{a}}-\frac{1}{a\binom{n+a+1}{a}}=\frac{1}{a}\left(\frac{1}{\frac{(n+a)!}{n!a!}}-\frac{1}{\frac{(n+a+1)!}{(n+1)!a!}}\right)\\=\frac{1}{a}\left(\frac{n!a!}{(n+a)!}-\frac{(n+1)!a!}{(n+a+1)!}\right)=\frac{1}{a}\frac{n!a!a}{(n+a+1)!}\\=\frac{n!a!}{(n+a+1)!}=\frac{1}{\frac{(n+a+1)!}{n!a!}}=\frac{1}{(a+1)\frac{(n+a+1)!}{n!(a+1)!}}=\frac{1}{(a+1)\binom{n+a+1}{a+1}}.$$
So, by indunction on $a$, Leibovici's formula holds.

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$$\sum_{k=0}^{n} \frac{(-1)^k}{k^3+9k^2+26k+24}\binom{n}{k}=\frac{1}{2(n+3)(n+4)}$$ holds.

$$\frac{1}{k^3+9k^2+26k+24}=\frac{1}{2}\frac{1}{k+2}-\frac{1}{k+3}+\frac{1}{2}\frac{1}{k+4}$$ holds.
By Leibovici's formula, $$\sum_{k=0}^{n} \frac{(-1)^k}{k^3+9k^2+26k+24}\binom{n}{k}\\=\frac{1} {2}\sum_{k=0}^n (-1)^k\frac{\binom{n}{k}}{k+2}-\sum_{k=0}^n (-1)^k\frac{\binom{n}{k}}{k+3}+\frac{1}{2}\sum_{k=0}^n (-1)^k\frac{\binom{n}{k}}{k+4}\\= \frac{1}{2}\frac{1}{2\binom{n+2}{2}}-\frac{1}{3\binom{n+3}{3}}+\frac{1}{2}\frac{1}{4\binom{n+4}{4}}= \dots =\frac{1}{2(n+3)(n+4)}.$$