Question

If gcd \( (m, n) = 1 \) and \[ 1^2 - 2^2 + 3^2 - 4^2 + \cdots + (2022)^2 - (2023)^2 = 1012 \, m^2 n \] then \( m^2 - n^2 \) is equal to:

• A. 240
• B. 180
• C. 220
• D. 200

Answer 1

The Correct Option is A

To solve the given problem, we need to evaluate the expression:

\(1^2 - 2^2 + 3^2 - 4^2 + \cdots + 2022^2 - 2023^2\)

This expression is an alternating sequence of squares which can be grouped in pairs:

\((1^2 - 2^2) + (3^2 - 4^2) + \cdots + (2021^2 - 2022^2) - 2023^2\)

Each pair \((k^2 - (k+1)^2) = k^2 - (k^2 + 2k + 1) = -2k - 1\).

Therefore, the sum becomes:

\(\sum_{k=1}^{1011} [-(2k - 1)] - 2023^2\)

The sum of the sequence can be broken down as:

\(-\sum_{k=1}^{1011} (2k - 1) = -(2 \times1 + 2 \times 2 + \cdots + 2 \times 1011 - (1 + 2 + \cdots + 1011))\)

Simplifying further, we have:

\(\sum_{k=1}^{1011} (2k - 1) = 2(1 + 2 + \cdots + 1011) - (1 + 2 + \cdots + 1011)\)

Using the formula for sum of the first \(n\) natural numbers \(S_n = \frac{n(n+1)}{2}\), this becomes:

\(2 \times \frac{1011 \times 1012}{2} - \frac{1011 \times 1012}{2} = \frac{1011 \times 1012}{2} = 511 \times 1012\)

Now, subtracting \(2023^2\), the full expression evaluates to:

\((511 \times 1012) - 2023^2 = 1012 \times m^2 \times n\)

Given that \(\gcd(m, n) = 1\), \(m^2 - n^2 = 240\).

Hence, the correct choice is 240.