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JEE Main 20236 AprilEvening Shift
Exam
JEE Main
Year
2023
Date
6 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2023 · 6 April · Evening Shift · Mathematics · Sequences and Series

If (m, n) = 1 and 1 ^ 2 - 2 ^ 2 + 3 ^ 2 - 4 ^ 2 + . . . . + (2021)...

If gcd⁡(m,n)=1\gcd \left(m, n\right) = 1 and 12−22+32−42+....+(2021)2−(2022)2+(2023)2=1012m2n{1}^{2} - {2}^{2} + {3}^{2} - {4}^{2} + . . . . + {\left(2021\right)}^{2} - {\left(2022\right)}^{2} + {\left(2023\right)}^{2} = 1012{m}^{2}n then m2−n2{m}^{2} - {n}^{2} is equal to

Options

  1. A240240
  2. B200200
  3. C220220
  4. D180180

Correct answer

A. 240240

Step-by-step solution

Let

S=12−22+32−42+....+(2021)2−(2022)2+(2023)2S = {1}^{2} - {2}^{2} + {3}^{2} - {4}^{2} + . . . . + {\left(2021\right)}^{2} - {\left(2022\right)}^{2} + {\left(2023\right)}^{2}

⇒S=(1−2)(1+2)+(3−4)(3+4)+....+(2021−2022)(2021+2022)+(2023)2⇒ S = \left(1 - 2\right)\left(1 + 2\right) + \left(3 - 4\right)\left(3 + 4\right) + . . . . + \left(2021 - 2022\right)\left(2021 + 2022\right) + {\left(2023\right)}^{2}

⇒S=−[3+7+11+15+....+4043]+(2023)2⇒ S = - \left[3 + 7 + 11 + 15 + . . . . + 4043\right] + {\left(2023\right)}^{2}

The number of terms in the bracket are 20222=1011\dfrac{2022}{2} = 1011⇒S=−10112(6+1010×4)+(2023)2⇒ S = - \dfrac{1011}{2}\left(6 + 1010 \times 4\right) + {\left(2023\right)}^{2}

⇒S=−1011×2023+(2023)2⇒ S = - 1011 \times 2023 + {\left(2023\right)}^{2}

⇒S=2023×1012⇒ S = 2023 \times 1012

⇒S=172×7×1012⇒ S = {17}^{2} \times 7 \times 1012

So,

m=17,n=7m = 17, n = 7 and gcd⁡(17,7)=1\gcd \left(17, 7\right) = 1

Hence, m2−n2=172−72=240{m}^{2} - {n}^{2} = {17}^{2} - {7}^{2} = 240

Hence this is the correct option.

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