Calculate 𝑇6, substitute it into 𝑁 = 2𝑇6 + 4, and verify the total by separating repeated and fixed parts.

Assuming $T_n$ refers to the $n$-th triangular number, the formula is $T_n = \frac{n(n+1)}{2}$.

$$T_6 = \frac{6(6+1)}{2}$$ $$T_6 = \frac{6 \times 7}{2}$$ $$T_6 = \frac{42}{2}$$ $$T_6 = 21$$

Substitute the calculated value of $T_6$ into the given equation for $N$.

$$N = 2(21) + 4$$ $$N = 42 + 4$$ $$N = 46$$

The expression for $N$ is $2T_6 + 4$. In this form, $2T_6$ represents the repeated part and 4 represents the fixed part.
The repeated part is $2 \times T_6 = 2 \times 21 = 42$.
The fixed part is 4.
Adding these parts together:

$$\text{Total} = \text{Repeated Part} + \text{Fixed Part}$$ $$\text{Total} = 42 + 4$$ $$\text{Total} = 46$$

This matches the value of $N$ calculated in Step 2.

The calculated value of $T_6$ is $\boxed{21}$.
The calculated value of $N$ is $\boxed{46}$.
The verification confirms the total is $\boxed{46}$.