query: Proof of equation with binomial coefficients: $\sum\limits_{k=1}^{n} (k+1) \binom{n}{k} = 2^{n-1} \cdot (n+2)-1$
positive_passages.docid: 1158
positive_passages.text: How to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$?
negative_passages.docid: 89576
negative_passages.text: Proof of a binomial identity $\sum_{k=0}^n {n \choose k}^{\!2} = {2n \choose n}.$
negative_passages.docid: 126001
negative_passages.text: Prove by Double Counting Method $\sum\limits_{k = 0}^m \binom{m}{k}\binom{n}{r + k} = \binom{m + n}{m + r}$
negative_passages.docid: 6539
negative_passages.text: Prove $\sum_{i=1}^n i! \cdot i = (n+1)! - 1$?
negative_passages.docid: 30338
negative_passages.text: How to prove that $\sum_{k=0}^n \binom nk k^2=2^{n-2}(n^2+n)$
negative_passages.docid: 6366
negative_passages.text: Combinatorial proof of summation of $\sum\limits_{k = 0}^n {n \choose k}^2= {2n \choose n}$
negative_passages.docid: 59638
negative_passages.text: How to prove this result on binomial coefficients?
negative_passages.docid: 108401
negative_passages.text: Sum of binomial coefficients $\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{2n - 2k}{n - 1} = 0$
negative_passages.docid: 136580
negative_passages.text: Proving $\sum_{k=0}^n\binom{2n}{2k} = 2^{2n-1}$
negative_passages.docid: 10875
negative_passages.text: proof of binomial identity involving double sum
negative_passages.docid: 79124
negative_passages.text: Proving $\binom {n-1}{r-1}=\sum_{k=0}^r(-1)^k\binom r k \binom{n+r-k-1}{r-k-1}$
negative_passages.docid: 158508
negative_passages.text: Combinatorial Proof for Binomial Identity: $\sum_{k = 0}^n \binom{k}{p} = \binom{n+1}{p+1}$
negative_passages.docid: 19571
negative_passages.text: Combinatorial proof for $\sum_{k = 0}^n \binom {r+k} k = \binom {r + n + 1} n$
negative_passages.docid: 7143
negative_passages.text: Identity for convolution of central binomial coefficients: $\sum\limits_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$
negative_passages.docid: 32791
negative_passages.text: Proving this binomial identity $\sum_{k=0}^n {n+k \choose k} \frac{1}{2^{k}}= 2^{n}$
negative_passages.docid: 25961
negative_passages.text: Prove: $\binom{n}{0}F_0+\binom{n}{1}F_1+\binom{n}{2}F_2+\cdots+\binom{n}{n}F_n=F_{2n}$
negative_passages.docid: 30414
negative_passages.text: Prove using combinatorics $\sum\limits_{k=0}^n 2^k \binom{n}{k} \binom{n-k}{\lfloor{\frac{n-k}{2}}\rfloor}=\binom{2n+1}{n}$