JEE Main 2023 (Online) 6th April Evening ShiftQuestion 1
If $\gcd(m,n)=1$ and $1^{2}-2^{2}+3^{2}-4^{2}+\ldots+(2021)^{2}-(2022)^{2}+(2023)^{2}=1012\,m^{2}n$, then $m^{2}-n^{2}$ is equal to
Sequence and Series · Standard Sums (Σ formulas)
Solution
Formulas used. (1) $(k+1)^2-k^2=2k+1=k+(k+1)$(2) $1+2+\cdots+n=\frac{n(n+1)}{2}$Step 1 - Group the terms. Keep $1^2$ and pair the rest as $(3^2-2^2)+(5^2-4^2)+\cdots+(2023^2-2022^2)$: $$1+(2+3)+(4+5)+\cdots+(2022+2023)=1+2+3+\cdots+2023 .$$ Step 2 - Sum. $$\frac{2023\cdot2024}{2}=1012\cdot2023 .$$ Step 3 - Match. $2023=7\cdot17^2$, so $1012\,m^2n=1012\cdot17^2\cdot7$ gives $m=17,\ n=7$ (coprime). Step 4. $$m^2-n^2=289-49=240 .$$ Answer: (A)