If $\operatorname{gcd}(m, n) = 1$ and $1^2 - 2^2 + 3^2 - 4^2 + \ldots + (2021)^2 - (2022)^2 + (2023)^2 = 1012 m^2 n$,then $m^2 - n^2$ is equal to

The sum $1+3+11+25+45+71+\ldots$ up to $20$ terms is equal to:

Let $\langle a_n \rangle$ be a sequence such that $a_1+a_2+\ldots+a_n = \frac{n^2+3n}{(n+1)(n+2)}$. If $28 \sum_{k=1}^{10} \frac{1}{a_k} = p_1 p_2 p_3 \ldots p_m$,where $p_1, p_2, \ldots, p_m$ are the first $m$ prime numbers,then $m$ is equal to

The sum of the first $n$ terms of the series $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \dots$ is

The sum of an infinite geometric series is $3$. $A$ series,which is formed by squaring its terms,also has a sum of $3$. The first series is

Let $S_n = \sum_{k=1}^n (-1)^{k-1} \cdot k^2$ for $n \geq 1$. Given that $S_{2n} = -n(2n+1)$ for $n = 1, 2, 3, \ldots$,then $S_{77} =$