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JEE Mains · Maths · STD 11 - 8. sequence and series

यदि \(\operatorname{gcd}(m, n)=1\) तथा \(1^2-2^2+3^2-4^2+\ldots \ldots\). \(+(2021)^2-(2022)^2+(2023)^2=1012 m^2 n\) है, तो \(\mathrm{m}^2-\mathrm{n}^2\) बराबर है

  1. A \(200\)
  2. B \(240\)
  3. C \(220\)
  4. D \(180\)
Verified Solution

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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