3
$\begingroup$

I saw this problem as an exercise in Combinatorial Identities :-

Prove that $${n+2 \choose 3}=1\cdot n + 2 \cdot (n - 1) + \ldots + n \cdot 1\,.$$

After giving some time to this, I think that it is quite similar to the identity :-

${n \choose k} = {n - 1 \choose k - 1} + {n - 1 \choose k}$

But I don't know how to prove this algebraically , anyone please help me with this.

(Note that I am still not sure whether we can use that identity or not , I can also guess we can use Vandermonde's Identity here) .

$\endgroup$
2

7 Answers 7

5
$\begingroup$

I suggest proving it combinatorially. $\binom{n+2}3$ is the number of $3$-element subsets of the set $[n+2]=\{1,2,\ldots,n+2\}$. We can classify those sets by their middle elements: let $\mathscr{A}_k$ be the family of all $3$-element subsets of $[n+2]$ of the form $\{j,k,\ell\}$, where $j<k<\ell$; clearly

$$\binom{n+2}3=\sum_k|\mathscr{A}_k|\;.$$

Now prove that $|\mathscr{A}_k|=(k-1)(n+2-k)$ and determine the range of possible values of $k$ to complete the proof.

$\endgroup$
1
  • $\begingroup$ thanks for the suggestion , but right now i am seeking an algebraical proof . $\endgroup$
    – Anonymous
    May 1, 2020 at 15:35
3
$\begingroup$

Both are equal to the number of ways of selecting three numbers from $1, \ldots, n+2$ : the first one by definition, the second by choosing the middle number, say $i+1$, and then by choosing one of the $i$ available "to the left" to be the smaller one, and one of the $(n+2)-(i+1)=n+1-i$ available "to the right" to be the bigger one.

$\endgroup$
2
$\begingroup$

I am sure this identity has been proven here. I can't find it. Note that $$\sum_{k=0}^m\,\binom{k}{r}\,\binom{m-k}{s}=\binom{m+1}{r+s+1}\tag{*}$$ for every integers $m,r,s$ with $0\leq r,s\leq m$. A combinatorial proof is to count the number of $(r+s+1)$-subsets of $\{0,1,2,\ldots,m\}$. Clearly, there are $\displaystyle\binom{m+1}{r+s+1}$ such subsets.

For $k=0,1,2,\ldots,m$, there are precisely $\displaystyle\binom{k}{r}\,\binom{m-k}{s}$ subsets of sizer $r+s+1$ such that $k$ is the $(r+1)$-st smallest element of these sets. This proves (*). Now, the OP's problem is when $m:=n+1$, $r:=1$, and $s:=1$.

An algebraic proof of (*) can be seen by considering $$f(x):=\sum_{k=r}^\infty\,\binom{k}{r}x^{k-r}(1+x)^{m-k}=(1+x)^{m-r}\,\sum_{k=r}^\infty\,\binom{k}{r}\,\left(\frac{x}{1+x}\right)^{k-r}\,.$$ Thus, $$\begin{align}f(x)&=(1+x)^{m-r}\,\sum_{k=0}^\infty\,\binom{k+r}{r}\,\left(\frac{x}{1+x}\right)^k \\&=(1+x)^{m-r}\,\left(1-\frac{x}{1+x}\right)^{-r-1}=(1+x)^{m+1}\,.\end{align}$$ For each integer $t\geq 0$, let $[x^t]\,g(x)$ denote the coefficient of $x^t$ in a polynomial $g(x)$. Then, $$\sum_{k=0}^m\,\binom{k}{r}\,\binom{m-k}{m-k-s}=[x^{m-r-s}]\,f(x)=[x^{m-r-s}]\,(1+x)^{m+1}\,.$$ Ergo, $$\sum_{k=0}^m\,\binom{k}{r}\,\binom{m-k}{s}=\sum_{k=0}^m\,\binom{k}{r}\,\binom{m-k}{m-k-s}=\binom{m+1}{m-r-s}=\binom{m+1}{r+s+1}\,.$$

Edit. I found a combinatorial proof of (*) in this old link. Analytic proofs of (*) are also given here. Algebraic proofs of (*) can be found here.

$\endgroup$
1
$\begingroup$

$$\sum_{k=1}^n k(n+1-k)=(n+1)\sum_{k=1}^nk-\sum_{k=1}^nk^2$$Now apply the identities$$\sum_{k=1}^nk=\frac12n(n+1)\qquad\sum_{k=1}^nk^2=\frac16n(n+1)(2n+1)$$and simplify the result.

$\endgroup$
0
$\begingroup$

We are trying to show that $$1\!\cdot\!n \, + \, 2\!\cdot\!(n - 1)\, + \, \ldots \, + \, n\!\cdot\!1\ = {n+2 \choose 3}$$ Peter Foreman wrote the left hand side as $$ \sum_{k=1}^n k(n-k+1)$$ Another way to write the left hand side is $$\sum_{j=1}^n\sum_{k=1}^j k $$ The left hand side usually comes up as a summation nested within a summation. For example, on the first day of Christmas you get 1 gift; on the second day, you get $1+2$ gifts; on the 3rd day, $1+2+3$ gifts; .... $$\sum_{j=1}^{12} \sum_{k=1}^j k = \binom{12+2}{3} = 364$$

The Hockey Stick Identity is usually given as $$\sum_{i=r}^{m} \binom{i}{r} = \binom{m+1}{r+1}$$ But I have found this equivalent version useful because the summation starts at $k=1$ and goes to $n = m - r +1$: $$\sum_{k=1}^{n} \binom{k+r-1}{r} = \binom{n+r}{r+1}$$ For $r=1$ $$\sum_{k=1}^{n} \binom{k}{1} = \binom{n+1}{2}$$ For $r=2$ $$\sum_{j=1}^{n} \binom{j+1}{2} = \binom{n+2}{3}$$ So, \begin{align*} 1\!\cdot\!n \, + \, 2\!\cdot\!(n - 1)\, + \, \ldots \, + \, n\!\cdot\!1 = \sum_{j=1}^n\sum_{k=1}^j k\ &= \sum_{j=1}^n\sum_{k=1}^j \binom{k}{1}\\ &= \sum_{j=1}^n \binom{j+1}{2}\\ &= \binom{n+2}{3} \end{align*} This answers the question, but the pattern continues.

For $r=3$ $$\sum_{i=1}^{n} \binom{i+2}{3} = \binom{n+3}{4}$$ so $$\sum_{i=1}^{n}\sum_{j=1}^i\sum_{k=1}^j k\ = \binom{n+3}{4}$$

$\endgroup$
0
$\begingroup$

Use induction and your identity

$$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$$

Upon the inductive step, your RHS becomes

$$1(n+1) + 2(n) + \dots + n(2) + (n+1)1$$

$$= 1(n) + 2(n-1) + \dots + n(1) + \sum_{i=1}^{n+1} i$$

$$= \binom{n+2}{3} + \binom{n+2}{2}$$

$$= \binom{n+3}{3}$$

$\endgroup$
0
$\begingroup$

I think one must "emphasize" the steps in a proof by induction. I don't know how much I can do this for a reader. My try:

The satatement is the following: $$P(n): {n+2\choose3}=1.n+2.(n-1)+...+(n-1).2+n.1 \;\text{where}\;\; n\geq1.$$

For $n=1$, $P(1)$ is true, because ${1+2\choose3}=1.$

Assume that for $n$, the statement $P(n)$ is true.

We must prove the statement for $n+1$, that is, we must prove that $P(n+1)$ is true:

$$\large\begin{array} .{(n+1)+2\choose 3}&={n+3\choose 3}\\ &={n+2\choose 3}+{n+2\choose 2}\hspace{0.5cm}\text{(By Pascal's triangle main property)}\\ &=1.n+2.(n-1)+...+(n-1).2+n.1+\frac{(n+2)(n+1)}{2}\hspace{1cm}(*)\\ &=\sum_{i=1}^{n}i(n+1-i)+\sum_{i=1}^{n+1}i\hspace{1cm}(**)\\ &=\sum_{i=1}^{n}i(n+1-i)+\sum_{i=1}^{n}i+(n+1)\;\text{(Separating one term...)}\\ &=\sum_{i=1}^{n}(i(n+1-i)+i)+(n+1).1\;\text{(Combining the two sums...)}\\ &=\sum_{i=1}^{n}i(n+1-i+1)+(n+1).1\\ &=\sum_{i=1}^{n}i(n+2-i)+(n+1).1\\ &=1.(n+1)+2.n+...+(n-1).3+n.2+(n+1).1\\ \end{array}$$ Hence $P(n+1)$ is true. Therefore, $P(n)$ is true for all $n\geq 1$ by induction.

Explanation of $(*)$ and $(**)$: ${n+2\choose 2}=\frac{(n+2)(n+1)}{2}$ by definition of combination, but $\sum_{i=1}^{n+1}i=\frac{(n+2)(n+1)}{2}$ by Gauss sum.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .