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JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Analyzing +4 | -1
If gcd⁡(m,n)=1\operatorname{gcd}(m, n)=1 and 12−22+32−42+…..+(2021)2−(2022)2+(2023)2=1012m2n1^{2}-2^{2}+3^{2}-4^{2}+\ldots . .+(2021)^{2} -(2022)^{2}+(2023)^{2}=1012 m^{2} n then m2−n2m^{2}-n^{2} is equal to
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Correct option: A
240
The correct answer is A (240) because simplifying the series results in 2023 * 1012, which factors to (172 * 7) * (22 * 11 * 23). Since gcd(m,n) = 1, m = 17 and n = 7, so m2 - n2 = 240.

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