query: Formula $\sum_{k=0} ^n \binom{n}{k}\cdot k^2 $
positive_passages.docid: 10917
positive_passages.text: Give a combinatorial proof: $ n(n+1)2^{n-2} = \sum_{k=1}^{n}k^2\binom{n}{k} $
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: 120976
negative_passages.text: Proving $\sum_{k=0}^{n}k{n\choose k}^2 = n{2n-1 \choose n-1} $
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: 151085
negative_passages.text: Simplify the sum $ \sum_{i=0}^{k}(-1)^i i \binom{n}{i} \binom{n}{k-i}$
negative_passages.docid: 44111
negative_passages.text: Find the value of $\sum_{k=1}^{n}k\binom{n}{k}$?
negative_passages.docid: 68848
negative_passages.text: How to simplify $\sum_{k=1}^nk\cdot k!$
negative_passages.docid: 8806
negative_passages.text: Summation of binomial coefficients
negative_passages.docid: 60860
negative_passages.text: Combinatorial proof of $\sum_{i = 0}^{n} \binom{i}{r - 1} = \binom{n + 1}{r}$ and then use result to find a formula for $1^2 + 2^2 + \ldots + n^2$
negative_passages.docid: 7302
negative_passages.text: $\sum k! = 1! +2! +3! + \cdots + n!$ ,is there a generic formula for this?
negative_passages.docid: 111098
negative_passages.text: Prove $2^{2n} = \sum_{k=0}^{n}\binom{2n+1}{k}$
negative_passages.docid: 118372
negative_passages.text: Combination proof for $n(n+1)2^{n-2}=\sum_{k=1}^{n}k^2\binom{n}{k}$
negative_passages.docid: 18046
negative_passages.text: Prove the following equality: $\sum_{k=0}^n\binom {n-k }{k} = F_n$
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: 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: 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: 5332
negative_passages.text: Find the value of $\sum_0^n \binom{n}{k} (-1)^k \frac{1}{k+1}$