$$ \left( \color{blue}{2k+1}\right) \cdot n = 2kn+n $$

Remove the parentheses by changing the sign of each term within them.

$$ - \left( k-1 \right) = -k+1 $$

$$ \left( \color{blue}{2k+1-2}\right) \cdot n = 2kn+n-2n $$

Combine like terms:

$$ 2k+ \color{blue}{1} \color{blue}{-4} = 2k \color{blue}{-3} $$

Remove the parentheses by changing the sign of each term within them.

$$ - \left( k-1 \right) = -k+1 $$

Combine like terms:

$$ 2kn-k+ \color{blue}{1} + \color{blue}{1} = 2kn-k+ \color{blue}{2} $$

Combine like terms:

$$ 2kn+ \color{blue}{n} \color{blue}{-2n} = 2kn \color{blue}{-n} $$

Multiply each term of $ \left( \color{blue}{2kn-k+1}\right) $ by each term in $ \left( 2kn-k+2\right) $.

$$ \left( \color{blue}{2kn-k+1}\right) \cdot \left( 2kn-k+2\right) = 4k^2n^2-2k^2n+4kn-2k^2n+k^2-2k+2kn-k+2 $$

Combine like terms:

$$ 4k^2n^2 \color{blue}{-2k^2n} + \color{red}{4kn} \color{blue}{-2k^2n} +k^2 \color{green}{-2k} + \color{red}{2kn} \color{green}{-k} +2 = \\ = 4k^2n^2 \color{blue}{-4k^2n} +k^2+ \color{red}{6kn} \color{green}{-3k} +2 $$

Remove the parentheses by changing the sign of each term within them.

$$ - \left( 2k-3 \right) = -2k+3 $$

Multiply each term of $ \left( \color{blue}{2kn-k+1}\right) $ by each term in $ \left( 2kn-k+2\right) $.

$$ \left( \color{blue}{2kn-k+1}\right) \cdot \left( 2kn-k+2\right) = 4k^2n^2-2k^2n+4kn-2k^2n+k^2-2k+2kn-k+2 $$

Combine like terms:

$$ 4k^2n^2 \color{blue}{-2k^2n} + \color{red}{4kn} \color{blue}{-2k^2n} +k^2 \color{green}{-2k} + \color{red}{2kn} \color{green}{-k} +2 = \\ = 4k^2n^2 \color{blue}{-4k^2n} +k^2+ \color{red}{6kn} \color{green}{-3k} +2 $$

Multiply $2kn+n$ by $ \dfrac{2kn-n-2k+3}{2} $ to get $ \dfrac{4k^2n^2-4k^2n+4kn-n^2+3n}{2} $.

Step 1: Write $ 2kn+n $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 2kn+n \cdot \frac{2kn-n-2k+3}{2} & \xlongequal{\text{Step 1}} \frac{2kn+n}{\color{red}{1}} \cdot \frac{2kn-n-2k+3}{2} \xlongequal{\text{Step 2}} \frac{ \left( 2kn+n \right) \cdot \left( 2kn-n-2k+3 \right) }{ 1 \cdot 2 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 4k^2n^2 -\cancel{2kn^2}-4k^2n+6kn+ \cancel{2kn^2}-n^2-2kn+3n }{ 2 } = \frac{4k^2n^2-4k^2n+4kn-n^2+3n}{2} \end{aligned} $$

Subtract $ \dfrac{4k^2n^2-4k^2n+4kn-n^2+3n}{2} $ from $ \dfrac{4k^2n^2-4k^2n+k^2+6kn-3k+2}{2} $ to get $ \dfrac{k^2+2kn+n^2-3k-3n+2}{2} $.

To subtract expressions with the same denominators, we subtract the numerators and write the result over the common denominator.

$$ \begin{aligned} \frac{4k^2n^2-4k^2n+k^2+6kn-3k+2}{2} - \frac{4k^2n^2-4k^2n+4kn-n^2+3n}{2} & = \frac{4k^2n^2-4k^2n+k^2+6kn-3k+2}{\color{blue}{2}} - \frac{4k^2n^2-4k^2n+4kn-n^2+3n}{\color{blue}{2}} = \\[1ex] &=\frac{ 4k^2n^2-4k^2n+k^2+6kn-3k+2 - \left( 4k^2n^2-4k^2n+4kn-n^2+3n \right) }{ \color{blue}{ 2 }}= \frac{k^2+2kn+n^2-3k-3n+2}{2} \end{aligned} $$