If gcd(m,n) = 1 andn 1^2 - 2^2 + 3^2 - 4^2 + ... . + (2021)^2n - (2022)^2 + (2023)^2 = 1012m^2n, then m^2 - n^2 isn equal to

(d) : We have,nn n (1^2-2^2)+(3^2-4^2)+ * s+\(2021)^2-(2022)^2\+(2023)^2=1012 m^2 n n => underbrace\-3-7-11- * s-4043\+(2023)^2=1012 m^2 nn nnnIt is an A.P.nNow, -4043=-3+(n-1)(-4) => n=1011nSo, -(3+7+ * s+4043)+(2023)^2=1012 m^2 nnn n => -(1011 / 2)[3+4043]+(2023)^2=1012 m^2 n n => -2023 x 1011+(2023)^2=1012 m^2 n n => 2023[2023-1011]=1012 m^2 n n => 2023 x 1012=1012 m^2 n => m^2 n=2023=7 x 17^2 n => m=17 and n=7 n Therefore m^2-n^2=17^2-7^2=289-49=240n n

If gcd(m,n) = 1 and 1^{2} - 2^{2} + 3^{2} - 4^{2} + .... | Sequence & Series (Mathematics)

(d) : We have,\n

\n\begin{aligned}\n& \left(1^2-2^2\right)+\left(3^2-4^2\right)+\cdots+\left\{(2021)^2-(2022)^2\right\}+(2023)^2=1012 m^2 n \\\n& \Rightarrow \underbrace{\{-3-7-11-\cdots-4043\}}+(2023)^2=1012 m^2 n\n\end{aligned}\n

\n\nIt is an A.P.\nNow,

-4043=-3+(n-1)(-4) \Rightarrow n=1011

\nSo,

-(3+7+\cdots+4043)+(2023)^2=1012 m^2 n

\n\begin{aligned}\n& \Rightarrow-\frac{1011}{2}[3+4043]+(2023)^2=1012 m^2 n \\\n& \Rightarrow-2023 \times 1011+(2023)^2=1012 m^2 n \\\n& \Rightarrow 2023[2023-1011]=1012 m^2 n \\\n& \Rightarrow 2023 \times 1012=1012 m^2 n \Rightarrow m^2 n=2023=7 \times 17^2 \\\n& \Rightarrow m=17 \text { and } n=7 \\\n& \therefore m^2-n^2=17^2-7^2=289-49=240\n\end{aligned}\n

Explanation

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