0
$\begingroup$

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!

$\endgroup$
3
  • $\begingroup$ Use the same idea again. This time use the formula for sum of first consecutive squares as well. $\endgroup$ Feb 6, 2017 at 5:09
  • $\begingroup$ The problem is that I have to prove that that is the formula for first consecutive squares :/ $\endgroup$
    – The Bosco
    Feb 6, 2017 at 5:11
  • $\begingroup$ This is not the formula for the first $x$ consecutive squares. $\endgroup$ Feb 6, 2017 at 5:18

2 Answers 2

0
$\begingroup$

$$\begin{align} \sum_{r=3}^x \binom {r-1}2=\binom x3\end{align}$$ 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, $$\begin{align} \sum_{r=a}^x \binom {r}a=\binom {x+1}{a+1}\end{align}$$

$\endgroup$
9
  • $\begingroup$ How do you go from your second line to the third line? $\endgroup$
    – The Bosco
    Feb 6, 2017 at 5:19
  • $\begingroup$ $\binom 22 =\frac {2\cdot 1}{1\cdot 2}=\frac 12 (2\cdot 1)$; etc $\endgroup$ Feb 6, 2017 at 5:27
  • $\begingroup$ God damn it, I didn't try to even use the factorial representation. Thank you so much. $\endgroup$
    – The Bosco
    Feb 6, 2017 at 5:28
  • $\begingroup$ You're welcome :) $\endgroup$ Feb 6, 2017 at 5:29
  • $\begingroup$ Does this answer your question? Or were you looking to derive the sum of squares? $\endgroup$ Feb 6, 2017 at 5:38
0
$\begingroup$

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 $\binom{n}{3}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.