If gcd (m, n)=1 and 1²-2²+3²-4²+… . .+(2021)²-(2022)²+(2023)²=1012 m² n then m²-n² is equal to :

Solution

Maths sir · 4 steps

Pair each odd square with the even square before it, and every pair collapses to a plain sum of two numbers.

1. Group the terms. Keep 1² alone and pair the rest as (3²-2²)+(5²-4²)+…+(2023²-2022²). Each pair is a difference of squares of consecutive numbers.

   (2k+1)²-(2k)² = (2k+1)+(2k)

2. Add everything up. So the pairs give 2+3+4+5+…+2022+2023, and with the leading 1 the whole sum is 1+2+…+2023 = (2023 × 2024)/2 = 2023 × 1012.
3. Match with the form. Now 1012 m² n = 1012 × 2023, so m² n = 2023 = 17² × 7. This gives m = 17 and n = 7, and gcd(17, 7) = 1. The other split, m = 1 and n = 2023, makes m² - n² negative, so it matches no option.
4. So. m² - n² = 289 - 49 = 240, which is option (c).

The idea to keep. For alternating squares of consecutive numbers, use a²-b²=(a-b)(a+b). Each pair turns into a+b, and the whole series becomes a simple sum of natural numbers.

Answer from the NTA final answer key. Solution by PrepWiser.