1
$\begingroup$

Let $n,m \in \mathbb{N}$. Prove the identity $$\sum^{n}_{k=0}\binom{m+k}{k} = \binom{n+m+1}{n}$$

This seems very similar to Vandermonde identity, which states that for nonnegative integers we have $\sum^{m}_{k=0}\binom{m}{k}\binom{n}{r-k} = \binom{m+n}{r}$. But, clearly this identity is somehow different from it. Any ideas?

$\endgroup$
2
$\begingroup$

We can write $\displaystyle \sum^{n}_{k=0}\binom{m+k}{k} = \sum^{n}_{k=0}\binom{m+k}{m} = \binom{m+0}{m}+\binom{m+1}{m}+........+\binom{m+n}{m}.$

Now Using Coefficient of $x^r$ in $(1+x)^{t} $ is $\displaystyle = \binom{t}{r}.$

So we can write above series as...

Coefficient of $x^m$ in $$\displaystyle \left[(1+x)^m+(1+x)^{m+1}+..........+(1+x)^{m+n}\right] = \frac{(1+x)^{m+n+1}-(1+x)^{m}}{(1+x)-1} = \frac{(1+x)^{m+n+1}-(1+x)^{m}}{x}$$

above we have used Sum of Geometric Progression.

So we get Coefficient of $x^{m+1}$ in $\displaystyle \left[(1+x)^{m+n+1}-(1+x)^{m}\right] = \binom{m+n+1}{m+1} = \binom{m+n+1}{n}.$

$\endgroup$
3
$\begingroup$

Let m $\in \mathbb{N}$ For $n=0$, we have $$ \dbinom{m}{0} = \dbinom{m+1}{0} = 1$$ By recurrence, let $n \in \mathbb{N}$ such that $${\sum_{k=0}^n\dbinom{m+k}{k}}=\dbinom{n+m+1}{n}$$ By using the hypothesis we have: $${\sum_{k=0}^{n+1}\dbinom{m+k}{k}}=\sum_{k=0}^n\dbinom{m+k}{k}+\dbinom{n+m+1}{n+1}= \dbinom{n+m+1}{n}+\dbinom{n+m+1}{n+1}$$ By using Pascal's Rule : $$\sum_{k=0}^{n+1}\dbinom{m+k}{k}=\dbinom{n+m+2}{n+1}$$ QED

$\endgroup$
  • $\begingroup$ This is a proof by induction, am I right? $\endgroup$ – user72151 Nov 11 '15 at 17:10
  • $\begingroup$ Yes, it's the easiest way to prove it. $\endgroup$ – Samsa Nov 15 '15 at 20:14
1
$\begingroup$

For $t=0,1,\cdots, n$, $\binom{m+t}{m}$ is the coefficient of $x^m$ in the expansion of $(1+x)^{m+t}$.

Hence,

$$\binom{m}{0}+\binom{m+1}{1}+\cdots +\binom{m+n}{n},$$ i.e. $$\binom{m}{m}+\binom{m+1}{m}+\cdots +\binom{m+n}{m}$$ is the coefficient of $x^m$ in the expansion of $$(1+x)^m+(1+x)^{m+1}+\cdots +(1+x)^{m+n}.$$

Here, we have $$(1+x)^m+(1+x)^{m+1}+\cdots +(1+x)^{m+n}$$$$=\frac{(1+x)^m((1+x)^{n+1}-1)}{x}=\frac{(1+x)^{m+n+1}}{x}-\frac{(1+x)^m}{x}.$$

Since there is no term of $x^m$ in $\frac{(1+x)^m}{x}$, what we want is the coefficient of $x^{m+1}$ in $(1+x)^{m+n+1}$, i.e. $\binom{m+n+1}{m+1}=\binom{m+n+1}{n}$.

$\endgroup$
1
$\begingroup$

Using the coefficient of operator $[x^n]$ to denote the coefficient $a_n$ of $x^n$ in a series $A(x)=\sum_{k=0}^{\infty}a_kx^k$ could also be sometimes convenient.

\begin{align*} \sum_{k=0}^n\binom{m+k}{k}&=\sum_{k=0}^n\binom{m+k}{m}\\ &=\sum_{k=0}^n[x^m](1+x)^{m+k}\\ &=[x^m](1+x)^m\sum_{k=0}^n(1+x)^k\\ &=[x^m](1+x)^m\frac{1-(1+x)^{n+1}}{1-(1+x)}\\ &=-[x^{m+1}](1+x)^m\left(1-(1+x)^{n+1}\right)\\ &=[x^{m+1}](1+x)^{m+n+1}\\ &=\binom{m+n+1}{n}\\ \end{align*}

$\endgroup$
0
$\begingroup$

We may prove it by bijection.

The left side counts the number of ways in which we can choose a pair $(k, X)$ where $k \in \{0, \ldots, n\}$ and $X$ is a subset of $m$ elements in $[m+k] = \{0, \ldots, m+k-1\}$.

The right side counts the number of ways in which we can choose a subset $Y$ of $m+1$ elements in $[m+n+1] = \{0,\ldots, m+n\}$.

By letting $m+k$ in the left correspond to the maximum element in $Y$, we get a bijection.

More precisely, for each $k$ and each $m$-subset $X$ of $[m+k]$, we associate $(k,X)$ with the set $Y = X \cup \{m+k\}$.

Conversely, for each $(m+1)$-subset $Y$ of $[m+n+1]$, we associate it with the pair $(k,X)$ given by $k = \max Y - m$ and $X = Y-\{m+k\}$.

You can check that these maps are well defined, and that one is the inverse of the other.

$\endgroup$
0
$\begingroup$

Using the Gosper's algorithm in Maxima: AntiDifference(binomial(m+k,k),k); $$ \binom{m+k}{k} = (k+1)\frac{\binom{m+k+1}{k+1}}{m+1} - k\frac{\binom{m+k}{k}}{m+1} $$ and the sum telescopes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy