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This question is from Book of Proof by Richard H. Hammack. This is exercise number 13 of chapter 3 section 6.

This is exercise I think is related to Pascal identity, but I'm struggling to understand it.

Michael Rozenberg
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Mauricio Vega
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    Hint: Notice that $\binom{2}{2} = \binom{3}{3}$, which allows you to use Pascal's identity. That said, this identity can be proved by induction, which hinges on Pascal's identity, or by a combinatorial proof. – N. F. Taussig Aug 30 '20 at 16:42
  • Hint: assume n$\geq$3 then partition$$ (nC_3) = ({n-1}C_3) + ({n-1}C_2) = ({n-2}C_3) + ({n-2}C_2) + ({n-1}C_2) =.....$$ and so on to get your answer – SAGNIK UPADHYAY Aug 30 '20 at 16:56

3 Answers3

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There is a more generalised form of that (we assume that $\binom{a}{b}:=0$ is $b>a$)

Identity: $$\sum_{t=k}^{n}\binom{t}{k}=\binom{n+1}{k+1}\quad\text{for $n\geq k$}$$

Proof using telescoping summation method: $$\sum_{t=k}^{n}\binom{t}{k}=\sum_{t=k}^{n}\left\{\binom{t+1}{k+1}-\binom{t}{k+1}\right\}\\=\sum_{t=k}^{n}\binom{t+1}{k+1}-\sum_{t=k}^{n}\binom{t}{k+1}\\=\sum_{t=k+1}^{n+1}\binom{t+1}{k+1}-\sum_{t=k}^{n}\binom{t}{k}\\=\binom{n+1}{k+1}-\underbrace{\binom{k}{k+1}}_{\text{$0$ by definition}}\\=\binom{n+1}{k+1}$$

This identity is known as Hockey-stick identity.

ShBh
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By using the Pascal's identity and the telescoping summation we obtain: $$\sum_{k=2}^{n-1}\binom{k}{2}=\sum_{k=2}^{n-1}\left(\binom{k+1}{3}-\binom{k}{3}\right)=\binom{n}{3}-\binom{2}{3}=\binom{n}{3}.$$

Michael Rozenberg
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$$\sum _{k=2}^{n-1} \binom{k}{2}=\binom{n}{3}\tag{1}$$ By induction

It is trivially true for $n = 3$

Now suppose that $(1)$ holds for $n$ and let's prove it for $(n+1)$

$$\sum _{k=2}^{n} \binom{k}{2}=\sum _{k=2}^{n-1} \binom{k}{2}+\binom{n}{2}=\binom{n}{3}+\binom{n}{2}=\binom{n+1}{3}$$ which proves that $(1)$ holds for any $n$.

ShBh
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Raffaele
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