JEE Main 20236 Apr, Evening shiftMathematicsMediumSequence and Series179× since 2002If gcd (m,n)=1\operatorname{gcd}~(\mathrm{m}, \mathrm{n})=1gcd (m,n)=1 and 12−22+32−42+…..+(2021)2−(2022)2+(2023)2=1012 m2n1^{2}-2^{2}+3^{2}-4^{2}+\ldots . .+(2021)^{2}-(2022)^{2}+(2023)^{2}=1012 ~m^{2} n12−22+32−42+…..+(2021)2−(2022)2+(2023)2=1012 m2n then m2−n2m^{2}-n^{2}m2−n2 is equal to :AOption A. 220BOption B. 200COption C. 240DOption D. 180CheckSkip · show solution
