To solve this problem, we need to compute the sum of the squares of the first ( n ) even natural numbers.

Approach

The key insight here is to recognize that the ( k )-th even natural number is ( 2k ). Therefore, the sum of the squares of the first ( n ) even natural numbers can be derived as follows:

1. The square of the ( k )-th even number is ( (2k)^2 = 4k^2 ).
2. The sum of these squares is ( 4 \times (1^2 + 2^2 + ... + n^2) ).
3. The sum of the squares of the first ( n ) natural numbers is given by ( \frac{n(n+1)(2n+1)}{6} ).
4. Multiplying this by 4 gives the desired result: ( \frac{2n(n+1)(2n+1)}{3} ).

This formula is efficient and runs in constant time ( O(1) ) since it involves a fixed number of arithmetic operations regardless of ( n ).

Solution Code

n = int(input())
result = (2 * n * (n + 1) * (2 * n + 1)) // 3
print(result)

Explanation

Let's break down the formula:

• ( 2n ): Accounts for the even number factor (since each term is ( 2k )).
• ( (n+1) ) and ( (2n+1) ): These are part of the sum of squares formula for natural numbers.
• Division by 3: Simplifies the product after factoring out the 4 from the square of even numbers.

This approach ensures that we compute the result efficiently and correctly for any positive integer ( n ). For example:

• If ( n = 3 ), the sum is ( 2^2 + 4^2 + 6^2 = 4 + 16 + 36 = 56 ). Using the formula: ( (234*7)/3 = 168/3 = 56 ), which is correct.

Thus, the solution is both optimal and easy to implement.

Answer: The code will output the sum of the squares of the first ( n ) even natural numbers using the derived formula. For the given input (though not specified here), it will compute the correct result. For example, if input is 3, output is 56. If input is 2, output is 20, etc. The final answer depends on the input ( n ), but the code correctly computes it.

The code as written will output the correct sum for any input n. For example, if n=5, the output is (2*5*6*11)/3 = (660)/3 = 220.

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