query: Combinatorics identity; proof of $\sum_{n≥0}\left[\binom{n}{m}(z^n)\right]=\frac{z^m}{(1-z)^{m+1}}$
positive_passages.docid: 19914
positive_passages.text: Closed form for $\sum_{n=0}^{\infty} \binom{n+k}{k} x^n$ (Negative Binomial Theorem)
negative_passages.docid: 58247
negative_passages.text: Prove $\sum_{k=0}^n \binom{n+k}{k}=\binom{2n+1}{n}$
negative_passages.docid: 120074
negative_passages.text: Prove by Combinatorial Argument that $\binom{n}{k}= \frac{n}{k} \binom{n-1}{k-1}$
negative_passages.docid: 1139
negative_passages.text: Combinatorial proof of $\sum_{j=0}^{k} \binom{n}{j} = \sum_{j=0}^k \binom{n-1-j}{k-j}2^j$
negative_passages.docid: 14239
negative_passages.text: How to show $\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}=2^{n}$
negative_passages.docid: 156566
negative_passages.text: Combinatorial proof of a binomial identity: $\sum_k \binom {2r} {2k-1}\binom{k-1}{s-1} = 2^{2r-2s+1}\binom{2r-s}{s-1}$
negative_passages.docid: 72639
negative_passages.text: Show that $\sum_{k=0}^n\binom{3n}{3k}=\frac{8^n+2(-1)^n}{3}$
negative_passages.docid: 40784
negative_passages.text: $\sum_{k=m}^{n}(-1)^k\binom{n}{k}\binom{k}{m}=0, n>m\geq 0$
negative_passages.docid: 5049
negative_passages.text: Proving $\sum_{k=0}^n(-1)^k\binom nk=0$
negative_passages.docid: 63510
negative_passages.text: Prove: $\binom{n}{k}^{-1}=(n+1)\int_{0}^{1}x^{k}(1-x)^{n-k}dx$ for $0 \leq k \leq n$
negative_passages.docid: 42763
negative_passages.text: How to prove Vandermonde's Identity: $\sum_{k=0}^{n}\binom{R}{k}\binom{M}{n-k}=\binom{R+M}{n}$?
negative_passages.docid: 107397
negative_passages.text: Algebraic proof for the identity $\sum_{k=0}^m(-1)^{m-k}{{n \choose {k}}}= {{n-1}\choose m}$
negative_passages.docid: 133181
negative_passages.text: Hint on how to prove the identity $\sum_{0 \leq k \leq n } \binom{k}{m} = \binom{n + 1}{m + 1}$
negative_passages.docid: 67650
negative_passages.text: Short and intuitive proof that $\left(\frac{n}{k}\right)^k \leq \binom{n}{k}$
negative_passages.docid: 44928
negative_passages.text: Show $\sum_n \frac{z^{2^n}}{1-z^{2^{n+1}}} = \frac{z}{1-z}$
negative_passages.docid: 61483
negative_passages.text: Prove the following identity combinatorially
negative_passages.docid: 106008
negative_passages.text: Combinatorial Identity $(n-r) \binom{n+r-1}{r} \binom{n}{r} = n \binom{n+r-1}{2r} \binom{2r}{r}$