Show for $n$, $k$ $\in$ N, such that 1 $\leq k \leq n$, $\sum_{k = 1}^{n}\binom{n}{k}\binom{n}{k-1} = \frac{1}{2}\binom{2n+2}{n+1} - \binom{2n}{n}$

Question

We are given a hint to show that both sides are equal to $\binom{2n}{n+1}$

For the right side, I have computed that

$$\frac{1}{2}\binom{2n+2}{n+1} - \binom{2n}{n}$$ $$=> \frac{(2n+2)!}{2((n+1)!)^2} - \frac{(2n)!}{n!^2}$$ $$=> \frac{(2n)!}{n!^2} * \left [ \frac{(2n+2)(2n+1)}{2(n+1)(n+1)} - 1\right ]$$ $$=> \frac{2n!}{(n!)^2}\left [ \frac{2n+1}{n+1} - \frac{n+1}{n+1}\right ]$$ $$=> \frac{2n!}{(n!)^2} * \frac{n}{n+1}$$ $$=> \frac{2n!}{n(n-1)!n!} * \frac{n}{n+1}$$ $$=> \frac{2n!}{n!(n+1)(n-1)!}$$ $$=> \frac{2n!}{(n-1)!(n+1)!}$$ $$=> \binom{2n}{n+1}$$

However, the left side is really giving me some trouble. I am not sure how to deal with the summation. Any help would be appreciated.

See https://math.stackexchange.com/questions/645505/show-sum-i-1n-binomni-binomni-1-binom2nn-1?rq=1 — Kolja, Nov 19 '22 at 20:56
For what it's worth, I posted an answer and then deleted it. I deleted it because the analysis is virtually the same as one of the answers given in the MathSE problem linked in the comment of Kolja. — user2661923, Nov 19 '22 at 21:20
Does this answer your question? Combinatorial proof of a binomial coefficient summation: $\sum_{k=1}^n \binom nk \binom n{k-1} = \frac12\binom{2n+2}{n+1} - \binom{2n}n$ - found using an Approach0 search. Note the RHS can also be simplified using ... — John Omielan, Nov 19 '22 at 21:53
(cont.) Pascal's rule, along with $\binom{2n}{n+1}=\binom{2n}{2n-(n+1)}=\binom{2n}{n-1}$. This also then aligns the problem with Show $\sum_{i=1}^{n}\binom{n}{i}\binom{n}{i-1}=\binom{2n}{n-1}$ referenced in Kolja's comment. Also, there are other duplicates, e.g., Proof of $\sum_{k=1}^n \binom{n}{k} \binom{n}{n+1-k} = \binom{2n}{n+1}$ via induction. — John Omielan, Nov 19 '22 at 21:56
k is the index of summation so it should not be in the "Show for" part. — marty cohen, Nov 19 '22 at 22:40

score 3 · Answer 1 · answered Nov 19 '22 at 20:50

3

We have

$$\sum_{k=1}^n {n\choose k} {n\choose n+1-k} = [z^{n+1}] (1+z)^n \sum_{k=1}^n {n\choose k} z^k \\ = [z^{n+1}] (1+z)^n \sum_{k=0}^n {n\choose k} z^k = [z^{n+1}] (1+z)^n (1+z)^n = [z^{n+1}] (1+z)^{2n} \\ = {2n\choose n+1}.$$

answered Nov 19 '22 at 20:50

Marko Riedel

61,317

Show for $n$, $k$ $\in$ N, such that 1 $\leq k \leq n$, $\sum_{k = 1}^{n}\binom{n}{k}\binom{n}{k-1} = \frac{1}{2}\binom{2n+2}{n+1} - \binom{2n}{n}$

1 Answers1