# Sum of binomial coefficients $\sum_{k=n}^{r}\binom{k}{n}$

Is is possible to find the sum of a binomial coefficient series like: $\sum_{k=n}^{r}\dbinom{k}{n}$? Just a random thought.

• If you have meant $\displaystyle\sum_{r=1}^n r\binom nr$ $$\displaystyle r\binom nr=r\dfrac{n\cdot(n-1)!}{r\cdot(r-1)!\{n-1-(r-1)\}!}=n\binom{n-1}{r-1}$$ – lab bhattacharjee Apr 7 '15 at 18:57
• ${k \choose n} = 0$ if $k < n$, $1$ if $k=n$. Maybe you'd rather do $$\sum_{k=n}^r {k \choose n}$$ – Robert Israel Apr 7 '15 at 19:15
• yeah yeah im so sorry... i didnt notice the flaw in my question – user220382 Apr 7 '15 at 19:18

$$\sum_{k=n}^r {k \choose n} = \dfrac{r+1-n}{n+1} {r+1 \choose n}$$ Prove by induction on $r$.
EDIT: Actually this can be written as $${r+1 \choose n+1}$$ in which form it has a nice combinatorial interpretation. Suppose you want to have $r+1$ objects numbered $1$ to $r+1$, and you want to choose $n+1$ of them. How many ways to do it? Divide into cases according to the highest-numbered object chosen. If the highest-numbered object chosen has number $k+1$, the remaining $n$ objects can be chosen from $1,\ldots, k$ in ${k \choose n}$ ways, where of course we need $k \ge n$. Therefore $${{r+1} \choose {n+1}} = \sum_{k=n}^r {k \choose n}$$