query: Prove $\sum _{k=2}^{n} k(k-1) {n \choose k}=n(n-1)2^{n-2}$
positive_passages.docid: 142666
positive_passages.text: Algebric proof for the identity $n(n-1)2^{n-2}=\sum_{k=1}^n {k(k-1) {n \choose k}}$
negative_passages.docid: 44161
negative_passages.text: Using induction to prove $\sum_{k=0}^{N-1} 2^k = 2^N-1$
negative_passages.docid: 24588
negative_passages.text: Proving the combinatorial identity ${n \choose k} = {n-2\choose k-2} + 2{n-2\choose k-1} + {n-2\choose k}$
negative_passages.docid: 13236
negative_passages.text: Proving equality - a sum including binomial coefficient $\sum_{k=1}^{n}k{n \choose k}2^{n-k}=n3^{n-1}$
negative_passages.docid: 148442
negative_passages.text: Show that the sum $\sum_{k = 0}^n 2^k \binom{n}{k}$ is equal to $3^n$
negative_passages.docid: 70827
negative_passages.text: Prove that $\sum_{i = n - b}^{a - m} {a - i \choose m}{b + i \choose n} = {a + b + 1 \choose m + n + 1}$
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: 87941
negative_passages.text: Prove by mathematical induction that $\forall n \in \mathbb{N} : \sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k} = \sum_{k=n+1}^{2n} \frac{1}{k} $
negative_passages.docid: 92784
negative_passages.text: Intuitive (combinatorial) proof of $2^n=\sum_{k=0}^n {n\choose k}$
negative_passages.docid: 96961
negative_passages.text: A proof for the identity $\sum_{n=r}^{\infty} {n \choose r}^{-1} = \frac{r}{r-1}$
negative_passages.docid: 15164
negative_passages.text: Proof of $\sum^{2N}_{n=1} \frac{(-1)^{n-1}}{n} = \sum^{N}_{n=1} \frac{1}{N+n}$
negative_passages.docid: 69943
negative_passages.text: why is ${n+1\choose k} = {n\choose k} + {n\choose k-1}$?
negative_passages.docid: 44328
negative_passages.text: Prove that $\sum_{k=1} ^ n k^{-1} = \ln(n) + O(1)$
negative_passages.docid: 4991
negative_passages.text: Proving by induction that $ \sum_{k=0}^n{n \choose k} = 2^n$
negative_passages.docid: 17714
negative_passages.text: Proof $\sum_{k=1}^n (-1)^{k+1} \binom{n}{k}\frac{1}{k} = H_n$ by induction
negative_passages.docid: 19581
negative_passages.text: $\sum_{k=0}^n (-1)^k \binom{n}{k}^2$ and $\sum_{k=0}^n k \binom{n}{k}^2$
negative_passages.docid: 7083
negative_passages.text: Alternating sum of binomial coefficients: given $n \in \mathbb N$, prove $\sum^n_{k=0}(-1)^k {n \choose k} = 0$