Summation of binomial coefficients

Question

Is there a closed formula for:

$\sum_{i=1}^{N}{\binom{i+k}{i}}$

( k is a constant whole number )

See $\sum_{i=0}^n\binom{i+k-1}{k-1}=\binom{n+k}{k}$ – Bart Michels Apr 08 '15 at 15:06 — Bart Michels, Apr 08 '15 at 15:06

score 5 · Answer 1 · answered Apr 03 '15 at 18:38

5

Yes:

$$\sum_{i=1}^N\binom{i+k}i=\sum_{i=1}^N\binom{i+k}k=\sum_{i=0}^N\binom{i+k}k-1=\binom{N+k+1}{k+1}-1\;.$$

answered Apr 03 '15 at 18:38

Brian M. Scott

That also is ${N + k + 1 \choose N} - 1$ (well that's basically just b.c. symetry, but I find this one more elegant :3) – servabat Apr 03 '15 at 18:45
@servabat: It is indeed. – Brian M. Scott Apr 03 '15 at 18:47
Can you please tell me how we get $\sum_{i=0}^N\binom{i+k}k-1=\binom{N+k+1}{k+1}-1$ ? – rayu Apr 03 '15 at 18:54
@Rayu: Ignoring the $-1$ on both sides, it’s a very standard identity involving binomial coefficients. See this question and answer for a couple of proofs. – Brian M. Scott Apr 03 '15 at 18:55
@Rayu: You’re welcome. – Brian M. Scott Apr 03 '15 at 18:58

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