The problem asks us to calculate the value of:

$2023^2 - 2020^2 + 2017^2 - 2014^2 + \dots + 7^2 - 4^2 + 1^2$

Step 1: Grouping terms using the difference of squares formula

The formula for the difference of squares is:

$a^2 - b^2 = (a - b)(a + b)$

Using this, each pair of terms like $2023^2 - 2020^2$, $2017^2 - 2014^2$, etc., can be rewritten.

For $2023^2 - 2020^2$:

$2023^2 - 2020^2 = (2023 - 2020)(2023 + 2020) = 3 \cdot 4043$

For $2017^2 - 2014^2$:

$2017^2 - 2014^2 = (2017 - 2014)(2017 + 2014) = 3 \cdot 4031$

Similarly for all pairs:

Each term follows the same structure, with the difference $a - b = 3$, and the sum decreasing by 12 each time (e.g., $4043, 4031, 4019, \dots$).

Step 2: Sum of terms

The sequence formed by the sums $4043, 4031, 4019, \dots$ is an arithmetic progression: $a_n = 4043 - 12(n-1)$ where $a_1 = 4043$, and the common difference $d = -12$.

To find the total number of terms ($n$), note that the sequence ends at $7^2 - 4^2 = (7 - 4)(7 + 4) = 3 \cdot 11$. Solving: $4043 - 12(n-1) = 11$ $12(n-1) = 4043 - 11 = 4032$ $n-1 = 336 \quad \Rightarrow \quad n = 337$

Step 3: Sum of the arithmetic progression

The sum of an arithmetic progression is given by: $S_n = \frac{n}{2} \cdot (a_1 + a_n)$ Here, $n = 337$, $a_1 = 4043$, and $a_n = 11$: $S_{337} = \frac{337}{2} \cdot (4043 + 11) = \frac{337}{2} \cdot 4054 = 337 \cdot 2027 = 683699$

Step 4: Adding the last term

The last term $+1^2 = 1$: $683699 + 1 = 683700$

Final Answer:

$\boxed{683700}$

Would you like additional explanations or details? Let me know!

Related Questions:

1. How is the difference of squares formula derived, and how is it applied here?
2. Can you explain how to determine the common difference in an arithmetic sequence?
3. How do you derive the formula for the sum of an arithmetic progression?
4. What is the role of $1^2$ in the final calculation, and why is it not paired with any term?
5. What are other scenarios where the difference of squares is useful?

Tip:

Always look for patterns in sequences and symmetry in problems to simplify complex calculations!