query: The sum of the first $n$ squares $(1 + 4 + 9 + \cdots + n^2)$ is $\frac{n(n+1)(2n+1)}{6}$
positive_passages.docid: 16472
positive_passages.text: Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$
negative_passages.docid: 59252
negative_passages.text: Show $\sum_{k=0}^{m} \binom{n-k}{m-k}=\binom{n+1}{m}$
negative_passages.docid: 129801
negative_passages.text: Finding a closed form for $\sum^{n}_{k=1} \frac{k}{(k+1)!} $
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: 57549
negative_passages.text: Finding the sum $\sum_{k=1}^rk^2\binom {n-k}{r-k}$
negative_passages.docid: 58264
negative_passages.text: How to derive the formula for the sum of the first $n$ perfect squares?
negative_passages.docid: 37007
negative_passages.text: Evaluating $\sum_{k=1}^{n} \frac{1}{k(k+1)}$
negative_passages.docid: 20271
negative_passages.text: Proving $\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}$ without induction
negative_passages.docid: 48200
negative_passages.text: Simplify $\sum_{i=0}^n (i+1)\binom ni$
negative_passages.docid: 50972
negative_passages.text: Prove that $\sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}$
negative_passages.docid: 103840
negative_passages.text: Calculate $\sum_{k=1}^n k2^{n-k}$
negative_passages.docid: 74512
negative_passages.text: How to show $n(n+1)(2n+1) \equiv 0 \pmod 6$?
negative_passages.docid: 9435
negative_passages.text: Prove that $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ for $n \in \mathbb{N}$.
negative_passages.docid: 37059
negative_passages.text: limit of the sum $\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} $
negative_passages.docid: 54355
negative_passages.text: $1+x+\ldots+x^n$ perfect square
negative_passages.docid: 10920
negative_passages.text: $\delta_2(n)$ the sum of the squares of the positive divisors of $n$.
negative_passages.docid: 73191
negative_passages.text: Proving that $\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3 \cdot 4} +\ldots + \frac{1}{n(n+1)} = \frac{n}{n+1}$