If operatorname{gcd}(m, n) = 1 and 1^2 - 2^2 | vedclass

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(B) The given expression is $S = (1^2 - 2^2) + (3^2 - 4^2) + \ldots + (2021^2 - 2022^2) + 2023^2$. Using the identity $a^2 - b^2 = (a - b)(a + b)$,we

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$240$

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(B) The given expression is $S = (1^2 - 2^2) + (3^2 - 4^2) + \ldots + (2021^2 - 2022^2) + 2023^2$. Using the identity $a^2 - b^2 = (a - b)(a + b)$,we have: $S = (1 - 2)(1 + 2) + (3 - 4)(3 + 4) + \ldots + (2021 - 2022)(2021 + 2022) + 2023^2$. $S = -1(1 + 2 + 3 + 4 + \ldots + 2022) + 2023^2$. The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$,so: $S = -\frac{2022 	imes 2023}{2} + 2023^2$. $S = -1011 	imes 2023 + 2023^2$. $S = 2023(2023 - 1011) = 2023 	imes 1012$. Given $1012 m^2 n = 2023 	imes 1012$,we get $m^2 n = 2023$. Since $2023 = 17^2 	imes 7$,we have $m = 17$ and $n = 7$. Thus,$m^2 - n^2 = 17^2 - 7^2 = 289 - 49 = 240$.

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

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