0

In my book has the follow :

For all $n \in\mathbb{N^*}$ and $ i \in\mathbb{N}$ with $0 \le i \le n$. We have this:

$$\binom{n}{i} + \binom{n}{i+1} = \binom{n+1}{i+1}$$

For $i = n$, the book said is easy to proof, after I have asked in here, I understand because for to 1 this proof is simple, how like in the comments is because of the 0 $\le$ i $\le$ n.

Then, when the $i = n$:

\begin{align} \binom{n}{n} + \binom{n}{n+1} &= \frac{n!}{n!(n-n)!} + \frac{n!}{n+1!(n-(n+1))!}\\ = 1\end{align}

because i > n , is defined $\binom{n}{i} = 0.$

Ju Grigori
  • 385
  • $\binom{n}{n+1}$ is equal to $0$. The formula $\binom{n}{k}=n!/(k!(n-k)!)$ is only valid when $0\le k\le n$. – Mike Earnest Apr 20 '18 at 02:32
  • Thank you! I don't have saw these fact about this range 0 $\le$ k $\le$ n – Ju Grigori Apr 20 '18 at 02:46
  • No problem. The reason for the restriction is that when you try to apply this formula to $\binom{n}{n+1}$, you get $\frac{n!}{(n+1)!(-1)!}$, and $(-1)!$ is undefined. – Mike Earnest Apr 20 '18 at 02:48
  • In line 1 you have $(n-(n+1))! = (-1)!$ is undefined. The definition ${n \choose k} = \frac {n!}{(n-k)!k!}$ only holds for $0 \le k \le n$. If $k > n$ then ${n \choose k} = 0$. – fleablood Apr 20 '18 at 03:05

1 Answers1

1

First of all notice that $$\binom{n}{i} + \binom{n}{i+1} = $$

by definition $$ \frac {n!}{i!(n-i)!}+ \frac {n!}{(i+1)!(n-i-1)!} =$$

Common denominator $$ \frac {n!(i+1+n-i)}{(i+1)!(n-i)!}=$$

Simplify $$\frac {(n+1)!}{(i+1)!(n-i)!}=$$

by definition $$ \binom{n+1}{i+1}.$$

Mohammad Riazi-Kermani
  • 68,728