If given $\sum_{r=1}^{m-1}\binom r3$, how does the summation evaluate when $n<r$ in $\binom nr$?

Question

Correct me if I'm running the summation correctly -

$$\sum_{r=1}^{m-1}\binom r3=\binom 13+\sum_{r=2}^{m-1}\binom r3$$
$$\sum_{r=1}^{m-1}\binom r3=\binom 13+\binom 23+\sum_{r=3}^{m-1}\binom r3$$
$$\sum_{r=1}^{m-1}\binom r3=\binom 13+\binom 23+\binom 33+\sum_{r=4}^{m-1}\binom r3$$
......so on How do you solve $\binom 13$ or any $\binom nr$ where $n<r$?

Answer 1

When $n < r$,

$$\binom{n}{r}=0$$

Your sum is equivalent to

$$\sum\limits_{r=3}^{m-1}\binom{r}{3} = \sum\limits_{r=3}^{m-1}\frac{r (r-1)(r-2)}{6} = \sum\limits_{r=3}^{m-1}\frac{r^3-3r^2+2r}{6}$$

$$= \frac{1}{6}\sum\limits_{r=3}^{m-1} r^3 - \frac{1}{2}\sum\limits_{r=3}^{m-1} r^2+\frac{1}{3}\sum\limits_{r=3}^{m-1} r$$

$$= \left[\frac{1}{6}\sum\limits_{r=1}^{m-1} r^3 - \frac{1}{2}\sum\limits_{r=1}^{m-1} r^2+\frac{1}{3}\sum\limits_{r=1}^{m-1} r\right]-\left[\frac{1}{6}\sum\limits_{r=1}^{2} r^3 - \frac{1}{2}\sum\limits_{r=3}^{2} r^2+\frac{1}{3}\sum\limits_{r=3}^{2} r\right]$$

You can finish the problem from here using the formulas for the sum of the first $k$ integers, first $k$ squares, and first $k$ cubes.

Answer 2

Note that $$\sum_{r=0}^{n}\binom ra=\sum_{r=a}^n\binom ra=\sum_{r=a+1}^n\binom ra=\binom{n+1}{a+1}$$ as $$\binom ra=0\qquad \text{for} \qquad r<a$$

Hence, putting $n=m-1$ and $a=3$, we have $$\sum_{r=0}^{m-1}\binom r3=\sum_{r=3}^{m-1}\binom r3=\sum_{r=4}^{m-1}\binom r3=\binom m4\qquad \blacksquare$$