I can't work out how to prove this equation is true by proof of mathematical
Use mathematical induction to prove that, for $n \ge 3$
$$\sum_{j=3}^n \binom{j-1}{2} = \binom{n}{3}$$
Please help, thanks
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An easier combinatorial argument: Count how many three-element subsets there are of ${1,2,3,\dots,n}$ 1) Directly 2) By breaking into cases based on the largest element appearing. – JMoravitz Jul 20 '18 at 02:23
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Related: Hockey Stick Identity. – JMoravitz Jul 20 '18 at 02:24
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Possible duplicate of Proof of the Hockey-Stick Identity: $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$ – JMoravitz Jul 20 '18 at 02:24
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The original question did ask for you to prove something similar to Hockey-Stick identity. $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$ before this equation – Harry Michael Kirwan Jul 20 '18 at 03:21
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Some other related posts: Evaluate $\sum\limits_{k=1}^n k^2$ and $\sum\limits_{k=1}^n k(k+1)$ combinatorially and Simplify triangular sum of triangular numbers: $\sum_{i=1}^{n}(\frac12i(i+1))$. Also other posts linked to the latter might be of interest. – Martin Sleziak Jul 20 '18 at 08:08
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Assume $$\sum_{j=3}^n \binom{j-1}{2} = \binom{n}{3}$$ is true. Let's prove $$\sum_{j=3}^{n+1} \binom{j-1}{2} = \binom{n+1}{3}$$ So $$\sum_{j=3}^{n+1} \binom{j-1}{2} = \sum_{j=3}^{n} \binom{j-1}{2} + \binom{n}{2} = \binom{n}{3}+ \binom{n}{2} = \frac{n!}{3!(n-3)!}+\frac{n!}{2!(n-2)!} = \frac{(n+1)!}{3!(n-2)!} = \binom{n+1}{3}$$
Ahmad Bazzi
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You can't assume true for the original equation, I'm trying to prove that the original equation is true. – Harry Michael Kirwan Jul 20 '18 at 03:18
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1@HarryMichaelKirwan that's how induction works. You assume that it is true for $n$ and prove that it being true for $n$, implies it is true for $n+1$. You must also prove that it is true for smallest possible value of $n$. In this case that is $n=3$. – Sonal_sqrt Jul 20 '18 at 03:28
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@HarryMichaelKirwan, I could have told you it is true for $n-1$, then i'd do the same steps to prove it for $n$. Try it yourself !!! – Ahmad Bazzi Jul 20 '18 at 03:31
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@AhmadBazzi actually yeah my bad sorry, but what I don't understand is where you got the $\binom{n}{2}$ going from $\sum_{j=3}^{n+1} \binom{j-1}{2} $ – Harry Michael Kirwan Jul 20 '18 at 03:33
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@AhmadBazzi Yes I understand now... the ending term of the sum will be that. Thank you – Harry Michael Kirwan Jul 20 '18 at 03:53