query: Calculate $\sum_{k=0}^n k \binom{n}{k} p^k (1-p)^{n-k}$
positive_passages.docid: 4931
positive_passages.text: variation of the binomial theorem
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: 130223
negative_passages.text: Proving a special case of the binomial theorem: $\sum^{k}_{m=0}\binom{k}{m} = 2^k$
negative_passages.docid: 158774
negative_passages.text: Series of binomial coefficients: $\sum\limits_{n=k}^{\infty}{\binom nk}x^n=\frac{x^k}{(1-x)^{k+1}}$
negative_passages.docid: 94060
negative_passages.text: Sum of square binomial coefficients
negative_passages.docid: 70248
negative_passages.text: Closed-form expression for $\sum_{k=0}^n\binom{n}kk^p$ for integers $n,\,p$
negative_passages.docid: 116491
negative_passages.text: Closed form for $\sum_{k=0}^{m} {\binom {m}{k}} a^{k} (b+ck)^N$
negative_passages.docid: 48200
negative_passages.text: Simplify $\sum_{i=0}^n (i+1)\binom ni$
negative_passages.docid: 113321
negative_passages.text: Hint proving this $\sum_{k=0}^{n}\binom{2n}{k}k=n2^{2n-1}$
negative_passages.docid: 65672
negative_passages.text: Proof of the summation $n!=\sum_{k=0}^n \binom{n}{k}(n-k+1)^n(-1)^k$?
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: 19868
negative_passages.text: Evaluate: $ \sum _ {k=1}^{n}{\frac{k}{n}\binom{n}{k}t^k(1-t)^{n-k}}$
negative_passages.docid: 87589
negative_passages.text: Evaluate $\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+...+\binom{n}{2k}$
negative_passages.docid: 44111
negative_passages.text: Find the value of $\sum_{k=1}^{n}k\binom{n}{k}$?
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: 30672
negative_passages.text: Show that $\sum\limits_{k=0}^n\binom{2n}{2k}^{\!2}-\sum\limits_{k=0}^{n-1}\binom{2n}{2k+1}^{\!2}=(-1)^n\binom{2n}{n}$
negative_passages.docid: 155769
negative_passages.text: Prove $\sum_{k= 0}^{n} k \binom{n}{k} = n \cdot 2^{n-1}$ using the binomial theorem for $n\geq1$