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The problem in the image asks to prove that for every positive integer \( k \), the expression \( k^4 + 2k^3 - k^2 - 2k \) is divisible by 12. Here's the step-by-step proof:

### Step 1: Simplify the expression
We start with the expression:
\[
f(k) = k^4 + 2k^3 - k^2 - 2k
\]
Our goal is to prove that \( f(k) \) is divisible by 12 for all positive integers \( k \).

### Step 2: Check divisibility by 3
We will first prove that \( f(k) \) is divisible by 3. To do this, we evaluate \( f(k) \) modulo 3 (i.e., the remainder when \( f(k) \) is divided by 3).

#### Case 1: \( k \equiv 0 \pmod{3} \)
If \( k \equiv 0 \pmod{3} \), then:
\[
f(k) = 0^4 + 2(0^3) - 0^2 - 2(0) = 0
\]
Thus, \( f(k) \equiv 0 \pmod{3} \).

#### Case 2: \( k \equiv 1 \pmod{3} \)
If \( k \equiv 1 \pmod{3} \), then:
\[
f(k) = 1^4 + 2(1^3) - 1^2 - 2(1) = 1 + 2 - 1 - 2 = 0
\]
Thus, \( f(k) \equiv 0 \pmod{3} \).

#### Case 3: \( k \equiv 2 \pmod{3} \)
If \( k \equiv 2 \pmod{3} \), then:
\[
f(k) = 2^4 + 2(2^3) - 2^2 - 2(2) = 16 + 16 - 4 - 4 = 24
\]
Since \( 24 \equiv 0 \pmod{3} \), we have \( f(k) \equiv 0 \pmod{3} \).

Thus, in all cases, \( f(k) \) is divisible by 3.

### Step 3: Check divisibility by 4
Next, we prove that \( f(k) \) is divisible by 4 by evaluating \( f(k) \) modulo 4.

#### Case 1: \( k \equiv 0 \pmod{4} \)
If \( k \equiv 0 \pmod{4} \), then:
\[
f(k) = 0^4 + 2(0

Udowodnij, że dla każdej dodatniej liczby naturalnej k liczba k^4 + 2k^3 - k^2 - 2k jest liczbą podzielną przez 12. Zapisz pełny tok rozumowania.

This problem demonstrates how to prove the divisibility of the expression k^4 + 2k^3 - k^2 - 2k by 12 for any positive integer k. Explore the use of algebra and modular arithmetic in this detailed step-by-step solution. Suitable for students in Grades 10-12.

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