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My problem is with $\binom{x}{3} = \frac{1}{6}(x-2)(x-1)x $

For $\binom{x}{2}$, what I did was to use the formula $\binom{n+1}{k+1}=\binom{0}{k}+\binom{1}{k}+...+\binom{n}{k}$ to get: $\binom{x}{2} = \binom{0}{1} + \binom{1}{1} + \binom{2}{1} +...\binom{x-1}{1} = 1 + 2 + 3 + 4 + ... (x-1) = \frac{(x-1)x}{2}$.

For $\binom{x}{1}$ it was even simple because I would get a sum of $1$'s.

I am stuck with $\binom{x}{3}$.

If I go with the same formula as above I get: $\binom{m}{3} = \binom{0}{2} +\binom{1}{2} + ... + \binom{m-2}{2}$

Let $S(N) = \frac{N(N+1)}{2}$, and I get: $\binom{m}{3} = S(1) + S(2) + S(3) + S(4) + ... + S(m-2)$

Thanks for any help!

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    Use the same idea again. This time use the formula for sum of first consecutive squares as well. – 2017-02-06
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    The problem is that I have to prove that that is the formula for first consecutive squares :/ – 2017-02-06
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    This is *not* the formula for the first $x$ consecutive squares. – 2017-02-06

2 Answers 2

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$$\begin{align} \sum_{r=3}^x \binom {r-1}2=\binom x3\end{align}$$ i.e. $$\binom 22+\binom32+\binom42+\cdots+\binom{x-1}2=\binom x3$$ or $$\frac 12\big(2\cdot 1+3\cdot 2+4\cdot 3+\cdots+(x-1)(x-2)\big)=\frac 16 x(x-1)(x-2)$$

NB: In general, $$\begin{align} \sum_{r=a}^x \binom {r}a=\binom {x+1}{a+1}\end{align}$$

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    How do you go from your second line to the third line? – 2017-02-06
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    $\binom 22 =\frac {2\cdot 1}{1\cdot 2}=\frac 12 (2\cdot 1)$; etc – 2017-02-06
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    God damn it, I didn't try to even use the factorial representation. Thank you so much. – 2017-02-06
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    You're welcome :) – 2017-02-06
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    Does this answer your question? Or were you looking to derive the sum of squares? – 2017-02-06
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    Actually, I am trying to derive sum of squares, cubes, etc. – 2017-02-06
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    From the way the question is phrased, it appears that you are only looking for the sum whose closed form is $\binom x3$ (which is provided in my solution above). – 2017-02-06
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53111/discussion-between-the-bosco-and-hypergeometric). – 2017-02-06
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    I think the answer provided above satisfies the question as phrased. However if your intended question is a different one then another question should be posted (as you have already done). – 2017-02-06
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What you want to prove is that $\binom{n}{3} = \binom{2}{2} + \binom{3}{2} + \binom{4}{2} + ... + \binom{n-1}{2}$.

Rewrite $\binom{2}{2}$ as $\binom{3}{3}$, since both are equal to $1$. Now, the expression becomes: $\binom{n}{3} = \binom{3}{3} + \binom{3}{2} + \binom{4}{2} + ... + \binom{n-1}{2}$.

Consider Pascal's identity: $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$.

Rewrite $\binom{3+1-1}{3} + \binom{3+1-1}{3-1}$ as $\binom{3+1}{3} = \binom{4}{3}$.

Now consider the original series again. To show that $\binom{n}{3} = \binom{4}{3} + \binom{4}{2} + ... + \binom{n-1}{2}$.

Similarly, we can combine the binomial term with $4$, and then $5$, and so on until we are left with $\binom{n}{3}$.