To solve this problem, we need to find the sum of all possible 4-digit numbers that can be formed using the digits 1, 2, 3, and 4, where repetition of digits is allowed.

Step-by-Step Solution:

1. Number of possible 4-digit numbers:
   Since repetition of digits is allowed, for each of the four digits (thousands, hundreds, tens, and units), we have 4 possible choices (1, 2, 3, or 4). Therefore, the total number of possible 4-digit numbers is: $4 \times 4 \times 4 \times 4 = 256$

2. Contribution of each digit to the sum:
   Let's examine how each digit (1, 2, 3, or 4) contributes to the total sum. Consider the thousands place first. Each digit will appear in the thousands place in exactly one-fourth of all numbers, because every digit has an equal probability of being in any position.

   Since there are 256 total numbers, each digit will appear in the thousands place $\frac{256}{4} = 64$ times. The same applies for the hundreds, tens, and units places.

3. Sum of digits for each place:
   The sum of the digits 1, 2, 3, and 4 is: $1 + 2 + 3 + 4 = 10$ Therefore, the contribution of each place to the total sum is calculated as follows:

   • Thousands place:
     Each of the digits 1, 2, 3, and 4 appears 64 times in the thousands place, and their sum is multiplied by 1000 (since it's the thousands place): $64 \times 10 \times 1000 = 640,000$

   • Hundreds place:
     Similarly, the contribution of the hundreds place is: $64 \times 10 \times 100 = 64,000$

   • Tens place:
     The contribution of the tens place is: $64 \times 10 \times 10 = 6,400$

   • Units place:
     The contribution of the units place is: $64 \times 10 \times 1 = 640$

4. Total sum:
   Now, we sum up the contributions from all four places: $640,000 + 64,000 + 6,400 + 640 = 711,040$

Final Answer:

The sum of all possible 4-digit numbers that can be formed using the digits 1, 2, 3, and 4 is $\boxed{711,040}$.

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Would you like more details or have any further questions?

Here are 5 related questions to expand on this concept:

1. How would the sum change if repetition was not allowed?
2. How does the total number of 4-digit numbers change if we introduce more digits?
3. What would be the sum of all 3-digit numbers formed using the digits 1, 2, and 3?
4. How would the sum change if we used the digits 0, 1, 2, and 3 instead?
5. What is the average of all the 4-digit numbers formed?

Tip: In problems involving sums of digits in different places, it helps to calculate the contribution of each place separately for clarity.