I was trying to find the number of diagonals of a nth polygon by deduction. Let $a_n$ be the number of the diagonals of thr nth polygon. Then I've got $$a_{n+1}=a_n+\binom{n}{3}$$ $$a_n=\sum_{i=3}^{n-1} \binom{i}{3}$$ But we simply know that $a_n=\binom{n}{4}$, where $a_n$ is the number of diagonals. So we get $\binom{n}{4}=\sum_{i=3}^{n-2} \binom{i}{3}$. My question is whether it is a trivial equation or it can be generalized like $\binom{n}{k}=\sum_{i=k-1}^{n-1} \binom{i}{k-1}$. If it can be, how to prove that?
Asked
Active
Viewed 118 times
0
-
$$\binom nk-\binom{n-1}k=?$$ https://math.stackexchange.com/questions/1125923/proving-pascals-identity – lab bhattacharjee Jan 15 '19 at 14:03
-
4This is the Hockey-stick_identity: see https://en.wikipedia.org/wiki/Hockey-stick_identity – Robert Z Jan 15 '19 at 14:03
-
Can be proved by induction. – gandalf61 Jan 15 '19 at 14:11
3 Answers
4
We have to choose four numbers from $[n]$. Assume the largest chosen number is $i+1\in[4..n]$. The other three numbers then are arbitrary three numbers in $[i]$.
Christian Blatter
- 232,144
3
You may transform the sum into a telescoping sum using the "binomial" fact
- $\binom{i}{3} = \binom{i+1}{4} - \binom{i}{4}$
Hence, \begin{eqnarray*} \sum_{i=3}^{n-1} \binom{i}{3} & = & \sum_{i=3}^{n-1} \left(\binom{i+1}{4} - \binom{i}{4} \right)\\ & \stackrel{\binom{3}{4} = 0}{=} & \binom{n-1+1}{4} - 0\\ & = & \binom{n}{4}\\ \end{eqnarray*}
trancelocation
- 33,349
- 1
- 21
- 46
3
Let $k=3$ so that $4=k+1$. The binomial coefficient $\binom n{k+1}$ counts the $k+1$-element subsets $S$ of the $n$-element set $\{0,1,\ldots,n-1\}$. Let $S$ be such a set and $i=\max(S)$. Then $k\leq i<n$, and $S=S'\cup\{i\}$ where $S'$ can be any $k$-element subset of the $i$-element set $\{0,1,\ldots,i-1\}$. So $$\binom n{k+1}=\sum_{i=k}^{n-1}\binom ik.$$
Marc van Leeuwen
- 119,547