Sum of the series $\sum_{k=0}^{r}(-1)^k.(k+1).(k+2).\binom{n}{r-k} $

Question

for $n>3$, The sum of the series $\displaystyle \sum_{k=0}^{r}(-1)^k.(k+1).(k+2).\binom{n}{r-k} = $

where $\displaystyle \binom{n}{r} = \frac{n!}{r!.(n-r)!}$

My try:: I have expand the expression

$\displaystyle 1.2.\binom{n}{r}-2.3.\binom{n}{r-1}+3.4.\binom{n}{r-2}+........+(-1)^r.(r+1).(r+2).\binom{n}{0}$

Now after that how can i calculate it

Thanks

Answer 1

We shall use the combinatorial identity

$$\sum_{j=0}^{k}{(-1)^j\binom{n}{j}}=(-1)^k\binom{n-1}{k}$$

This can be proven easily by induction, and there is also probably some combinatorial argument why it holds. We shall use the equivalent form

$$\sum_{j=0}^{k}{(-1)^{k-j}\binom{n}{j}}=\binom{n-1}{k}$$

Now $(r-k+1)(r-k+2)=k(k-1)-(2r+2)k+(r^2+3r+2)$, so

\begin{align} & \sum_{k=0}^{r}{(-1)^k(k+1)(k+2)\binom{n}{r-k}} \\ &=\sum_{k=0}^{r}{(-1)^{r-k}(r-k+1)(r-k+2)\binom{n}{k}} \\ & =\sum_{k=2}^{r}{(-1)^{r-k}k(k-1)\binom{n}{k}}-(2r+2)\sum_{k=1}^{r}{(-1)^{r-k}k\binom{n}{k}}+(r^2+3r+2)\sum_{k=0}^{r}{(-1)^{r-k}\binom{n}{k}} \end{align}

We have

\begin{align} \sum_{k=2}^{r}{(-1)^{r-k}k(k-1)\binom{n}{k}} & =\sum_{k=2}^{r}{(-1)^{(r-2)-(k-2)}n(n-1)\binom{n-2}{k-2}} \\ & =n(n-1)\sum_{k=0}^{r-2}{(-1)^{(r-2)-k}\binom{n-2}{k}} \\ & =n(n-1)\binom{n-3}{r-2} \\ & =\frac{r(r-1)(n-r)}{n-2}\binom{n}{r} \end{align}

\begin{align} \sum_{k=1}^{r}{(-1)^{r-k}k\binom{n}{k}} & =\sum_{k=1}^{r}{(-1)^{(r-1)-(k-1)}n\binom{n-1}{k-1}} \\ & =n\sum_{k=0}^{r-1}{(-1)^{(r-1)-k}\binom{n-1}{k}} \\ & =n\binom{n-2}{r-1} \\ & =\frac{r(n-r)}{n-1}\binom{n}{r} \end{align}

\begin{align} \sum_{k=0}^{r}{(-1)^{r-k}\binom{n}{k}} & =\binom{n-1}{r} \\ & =\frac{n-r}{n}\binom{n}{r} \end{align}

Thus

\begin{align} & \sum_{k=0}^{r}{(-1)^k(k+1)(k+2)\binom{n}{r-k}} \\ & =\frac{r(r-1)(n-r)}{n-2}\binom{n}{r}-(2r+2)\frac{r(n-r)}{n-1}\binom{n}{r}+(r^2+3r+2)\frac{n-r}{n}\binom{n}{r} \\ & =\binom{n}{r}\frac{(n-r)(r(r-1)n(n-1)-(2r+2)rn(n-2)+(r^2+3r+2)(n-1)(n-2))}{n(n-1)(n-2)} \\ & =\binom{n}{r}\frac{(n-r)(2r^2+(6-4n)r+(2n^2-6n+4))}{n(n-1)(n-2)} \end{align}

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\mbox{Note that}\quad\left. \sum_{k = 0}^{r}\pars{-1}^{k}\pars{k + 1}\pars{k + 2}{n \choose r - k} \right\vert_{\ n\ >\ 3} = \left.\partiald[2]{}{x}\sum_{k = 0}^{r}{n \choose r - k}x^{k + 2} \right\vert_{\ x\ =\ -1} \end{align}

Moreover,

\begin{align} \sum_{k = 0}^{r}{n \choose r - k}x^{k + 2} & = \bracks{z^{r}}\sum_{\ell = 0}^{\infty}z^{\ell} \sum_{k = 0}^{\ell}{n \choose \ell - k}x^{k + 2} = x^{2}\bracks{z^{r}}\sum_{k = 0}^{\infty}x^{k} \sum_{\ell = k}^{\infty}{n \choose \ell - k}z^{\ell} \\[5mm] & = x^{2}\bracks{z^{r}}\sum_{k = 0}^{\infty}x^{k} \sum_{\ell = 0}^{\infty}{n \choose \ell}z^{\ell + k} = x^{2}\bracks{z^{r}}\sum_{k = 0}^{\infty}\pars{zx}^{k} \sum_{\ell = 0}^{\infty}{n \choose \ell}z^{\ell} \\[5mm] & = x^{2}\bracks{z^{r}}\pars{1 - zx}^{-1}\pars{1 + z}^{n} \end{align} and $\ds{\lim_{x \to -1}\partiald[2]{}{x} \bracks{x^{2}\pars{1 - zx}^{-1}} = {2 \over \pars{1 + z}^{3}}}$ such that $\ds{\bracks{z^{r}}\bracks{{2 \over \pars{1 + z}^{3}}\,\pars{1 + z}^{n}} = 2{n - 3 \choose r}}$

---

Finally, $$\bbox[15px,#ffe,border:1px dotted navy]{\ds{\left. \sum_{k = 0}^{r}\pars{-1}^{k}\pars{k + 1}\pars{k + 2}{n \choose r - k} \right\vert_{\ n\ >\ 3} = 2{n - 3 \choose r}}} $$

Answer 3

Felix Marin's approach is nice and short, but I thought I'd add another approach that uses standard Binomial tools: Vandermonde's Identity, the symmetry of Pascal's Triangle, and $\binom{n}{k}=\frac nk\binom{n-1}{k-1}$.

$$ \hspace{-12pt}\begin{align} &\phantom{={}}\sum_{k=0}^r(-1)^k(k+1)(k+2)\binom{n}{r-k}\\ &=\sum_{k=0}^r(-1)^{r-k}(r-k+1)(r-k+2)\binom{n}{k}\tag{1a}\\ &=\sum_{k=0}^r(-1)^{r-k}[k(k-1)-2(r+1)k+(r+1)(r+2)]\binom{n}{k}\tag{1b}\\ &=\scriptsize n(n-1)\sum_{k=0}^r\binom{-1}{r-k}\binom{n-2}{k-2}-2(r+1)n\sum_{k=0}^r\binom{-1}{r-k}\binom{n-1}{k-1}+(r+1)(r+2)\sum_{k=0}^r\binom{-1}{r-k}\binom{n}{k}\tag{1c}\\ &=n(n-1)\binom{n-3}{r-2}-2(r+1)n\binom{n-2}{r-1}+(r+1)(r+2)\binom{n-1}{r}\tag{1d}\\ &=\left(n(n-1)\frac{r-1}{n-2}-2(r+1)n+(r+1)(r+2)\frac{n-1}r\right)\binom{n-2}{r-1}\tag{1e}\\ &=\frac{2(n-r-1)(n-r-2)}{r(n-2)}\binom{n-2}{n-r-1}\tag{1f}\\ &=\frac{2(n-3)}{r}\binom{n-4}{n-r-3}\tag{1g}\\ &=2\binom{n-3}{r}\tag{1h}\\ \end{align} $$ Explanation:
$\text{(1a):}$ substitute $k\mapsto r-k$
$\text{(1b):}$ $(r-k+1)(r-k+2)=k(k-1)-2(r+1)k+(r+1)(r+2)$
$\text{(1c):}$ $\binom{n}{k}=\frac{n(n-1)}{k(k-1)}\binom{n-2}{k-2}$ and $\binom{n}{k}=\frac nk\binom{n-1}{k-1}$
$\phantom{\text{(1c):}}$ $\binom{-1}{r-k}=(-1)^{r-k}[k\le r]$ (Iverson Brackets)
$\text{(1d):}$ Vandermonde's Identity
$\text{(1e):}$ $\binom{n-3}{r-2}=\frac{r-1}{n-2}\binom{n-2}{r-1}$ and $\binom{n-1}{r}=\frac{n-1}{r}\binom{n-2}{r-1}$
$\text{(1f):}$ simplify the rational function
$\phantom{\text{(1f):}}$ $\binom{n-2}{r-1}=\binom{n-2}{n-r-1}$ (symmetry of Pascal's Triangle)
$\text{(1g):}$ $\binom{n-2}{n-r-1}=\frac{(n-2)(n-3)}{(n-r-1)(n-r-2)}\binom{n-4}{n-r-3}$
$\text{(1h):}$ $\binom{n-4}{n-r-3}=\binom{n-4}{r-1}$ (symmetry of Pascal's Triangle)
$\phantom{\text{(1h):}}$ $\binom{n-3}{r}=\frac{n-3}r\binom{n-4}{r-1}$

Answer 4

Suppose we seek to evaluate $$\sum_{k=0}^r (-1)^k (k+2)(k+1) {n\choose r-k}.$$

Start from $${n\choose r-k} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{r-k+1}} (1+z)^n \; dz.$$

This yields the following expression for the sum $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \sum_{k=0}^r (-1)^k (k+2)(k+1) \frac{1}{z^{r-k+1}} (1+z)^n \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z^{r+1}} \sum_{k=0}^r (-1)^k (k+2)(k+1) z^k \; dz.$$

Observe that the defining integral of the binomial coefficient is zero when $k > r.$ This condition is precisely $r-k+1 < 1$ which makes the integrand into an entire function. Therefore we may extend the sum to infinity.

We thus have $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z^{r+1}} \sum_{k=0}^\infty (-1)^k (k+2)(k+1) z^k \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z^{r+1}} \left(\frac{1}{1+z}\right)'' \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z^{r+1}} \frac{2}{(1+z)^3} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{2(1+z)^{n-3}}{z^{r+1}} \; dz$$

Therefore the value of the sum is given by $$[z^r] 2 (1+z)^{n-3} = 2{n-3\choose r}.$$

This may be re-written as $$2{n\choose r} \frac{(n-r)(n-r-1)(n-r-2)}{n(n-1)(n-2)}$$ which is indeed the same as the earlier results, namely $${n\choose r} \frac{(n-r)(2r^2 + (6-4n)r + 2n^2-6n+4}{n(n-1)(n-2)}.$$

A trace as to when this method appeared on MSE and by whom starts at this MSE link.

Answer 5

Maxima, after loading Zeilberger, gives me:

$$ \sum_{0 \le k \le r} (-1)^k (k + 1) (k + 2) \binom{n}{r - k} = - \binom{n}{r} \frac{(r - n) (2 r^2 + (6 - 4 n) r + 2 n^2 - 6 n + 4)}{n^3 - 3 n^2 + 2 n} $$