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MCQMediumJEE Main 2023 · 6 April, Shift 2Sum of Series

Mathematics Question from JEE Main 2023 · 6 April, Shift 2

If gcd⁡(m,n)=1\gcd(m, n) = 1 and

12−22+32−42+⋯+(2021)2−(2022)2+(2023)2=1012 m2n1^2 - 2^2 + 3^2 - 4^2 + \cdots + (2021)^2 - (2022)^2 + (2023)^2 = 1012\,m^2 n

then m2−n2m^2 - n^2 is equal to:

  • A

    180180

  • B

    220220

  • C

    200200

  • D

    240240

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