query: How can this sum $\sum_{k=0}^a \binom{n}{k}$ be simplified?
positive_passages.docid: 18793
positive_passages.text: Partial sum of rows of Pascal's triangle
negative_passages.docid: 69498
negative_passages.text: What is $\sum_{r=0}^n \frac{(-1)^r}{\binom{n}{r}}$?
negative_passages.docid: 71797
negative_passages.text: Combinatorial argument for $\sum\limits_{k=i}^{n}\binom{n}{k}\binom{k}{i} = \binom{n}{i}2^{n-i}$
negative_passages.docid: 107071
negative_passages.text: Prove $\sum_{k= 0}^{n} k \binom{n}{k} = n \cdot 2^{n - 1}$ using the binomial theorem
negative_passages.docid: 28027
negative_passages.text: Find $\sum_{k=0}^{n}4^k \binom{n}{k}$
negative_passages.docid: 47052
negative_passages.text: How can I simplify this expression involving binomial coefficients?
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: 87479
negative_passages.text: Closed form for $\sum_{k=0}^{n-p}\binom{n}{k}\binom{n}{p+k}$
negative_passages.docid: 94784
negative_passages.text: $\sum_{k=-m}^{n} \binom{m+k}{r} \binom{n-k}{s} =\binom{m+n+1}{r+s+1}$ using Counting argument
negative_passages.docid: 44111
negative_passages.text: Find the value of $\sum_{k=1}^{n}k\binom{n}{k}$?
negative_passages.docid: 103840
negative_passages.text: Calculate $\sum_{k=1}^n k2^{n-k}$
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: 49066
negative_passages.text: Give the combinatorial proof of the identity $\sum\limits_{i=0}^{n} \binom{k-1+i}{k-1} = \binom{n+k}{k}$
negative_passages.docid: 92704
negative_passages.text: Prove that $\sum_{k=1}^{n}k\binom{n}{k}=n\cdot2^{n-1}$
negative_passages.docid: 345
negative_passages.text: Proving $ \sum _{k=0} ^m \binom nk \binom{n-k}{m-k} = 2^m \binom {n}{m}$.
negative_passages.docid: 83400
negative_passages.text: Sum of products of binomial coefficients: $ \sum_{l=0}^{n-j} \binom{M-1+l}{l} \binom{n-M-l}{n-j-l} = \binom{n}{j} $
negative_passages.docid: 23557
negative_passages.text: Algebraic Proof that $\sum\limits_{i=0}^n \binom{n}{i}=2^n$