To solve this, we will break down the problem step by step.

We are given the double summation:

$\sum_{i=3}^{3} \sum_{j=0}^{3} (j \cdot i + j)$

1. Inner summation: First, let's focus on the inner sum:

   $\sum_{j=0}^{3} (j \cdot i + j)$ This expression can be split into two separate sums:

   $\sum_{j=0}^{3} (j \cdot i) + \sum_{j=0}^{3} j$

   • The first sum is:

     $\sum_{j=0}^{3} (j \cdot i) = i \cdot \sum_{j=0}^{3} j = i \cdot (0 + 1 + 2 + 3) = i \cdot 6 = 6i$

   • The second sum is:

     $\sum_{j=0}^{3} j = 0 + 1 + 2 + 3 = 6$

   Therefore, the inner summation becomes:

   $6i + 6$

2. Outer summation: Now, for the outer sum, we sum this expression over $i$ from 3 to 3 (which only takes the value $i = 3$):

   $\sum_{i=3}^{3} (6i + 6)$

   When $i = 3$, the expression becomes:

   $6 \cdot 3 + 6 = 18 + 6 = 24$

Thus, the value of the entire summation is:

$\boxed{24}$

Would you like further explanation or help with other problems?

Here are some related questions to explore:

1. What would happen if the outer sum was from $i = 1$ to $i = 3$?
2. How can we generalize this formula for summations of this type?
3. Can you modify this summation to include a higher power of $j$?
4. What is the result if the bounds of both summations are changed to different integers?
5. How would this change if we replace $i$ with a constant value?

Tip: For double summations like this, break the problem into manageable parts—starting with the inner sum, then moving to the outer sum.