Given the sequence a, ab, aab, aabb, aaabb,aaabbb,…. Upto 2004 terms, the total number of times a's and b' s are used from 1 to 2004 terms are :

Answer

To solve the problem of finding the total number of times 'a's and 'b's are used in the sequence up to 2004 terms, we will analyze the pattern in the sequence.

### Step-by-Step Solution:

1. **Identify the Sequence:**
The sequence given is:
- 1st term: a
- 2nd term: ab
- 3rd term: aab
- 4th term: aabb
- 5th term: aaabb
- 6th term: aaabbb
- ...

Each term consists of 'a's followed by 'b's, where the number of 'a's and 'b's follows a specific pattern.

2. **Count the Total Terms:**
We are given that we need to consider up to 2004 terms.

3. **Determine the Count of 'a's:**
- The pattern for the number of 'a's in the sequence is:
- 1, 1, 2, 2, 3, 3, 4, 4, ..., n, n
- This means for every integer n, it appears twice in the sequence.
- The last integer n that appears in the sequence when we reach 2004 terms is:
- \( n = \frac{2004}{2} = 1002 \)
- Therefore, the total number of 'a's used can be calculated as:
- \( 1 + 1 + 2 + 2 + 3 + 3 + ... + 1002 + 1002 \)
- This can be simplified to:
- \( 2 \times (1 + 2 + 3 + ... + 1002) \)

4. **Calculate the Sum of Natural Numbers:**
- The sum of the first n natural numbers is given by the formula:
\[ S_n = \frac{n(n + 1)}{2} \]
- For n = 1002:
\[ S_{1002} = \frac{1002 \times 1003}{2} = 502503 \]
- Therefore, the total number of 'a's is:
\[ 2 \times 502503 = 1005006 \]

5. **Determine the Count of 'b's:**
- The pattern for the number of 'b's is:
- 0, 1, 1, 2, 2, 3, 3, ..., n-1, n
- The last integer n that appears in the sequence when we reach 2004 terms is still 1002.
- Therefore, the total number of 'b's used can be calculated as:
- \( 0 + 1 + 1 + 2 + 2 + 3 + 3 + ... + 1001 + 1002 \)
- This can be simplified to:
- \( (0 + 1 + 2 + ... + 1001) + (1 + 2 + ... + 1002) \)

6. **Calculate the Sum of 'b's:**
- The first part (0 to 1001) is:
\[ S_{1001} = \frac{1001 \times 1002}{2} = 501501 \]
- The second part (1 to 1002) is:
\[ S_{1002} = \frac{1002 \times 1003}{2} = 502503 \]
- Therefore, the total number of 'b's is:
\[ 501501 + 502503 = 1002004 \]

### Final Counts:
- Total number of 'a's = 1005006
- Total number of 'b's = 1002004