JEE Mains · Maths · STD 11 - 8. sequence and series
If \(\operatorname{gcd}( m , n )=1\) and \(1^2-2^2+3^2-4^2+\ldots \ldots\) \(+(2021)^2-(2022)^2+(2023)^2=1012 m ^2 n\), then \(m ^2- n ^2\) is equal to
- A \(200\)
- B \(240\)
- C \(220\)
- D \(180\)
Answer & Solution
Correct Answer
(B) \(240\)
Step-by-step Solution
Detailed explanation
\(1^2-2^2+3^2-4^2+\ldots(2021)^2-(2022)^2+(2023)^2=1012\) \(m ^2 n\) \(=(1-2)(1+2)+(3-4)(3+4)+\ldots+(2021-2022)\) \(+2022)+(2023)^2\) \(=(-1)(1+2+3+4+\ldots .+2022)+(2023)^2\) \(=(-1) \cdot \frac{(2022)(2023)}{2}+(2023)^2\) \(=2023(2023-1011)=2023 \times 1012\)…
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