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' and 'b' are used in the given sequence up to 2004 terms, we can break it down into a step-by-step solution.

### Step 1: Understand the Sequence
The sequence is as follows:
1. a
2. ab
3. aab
4. aabb
5. aaabb
6. aaabbb
7. ...

From the pattern, we can see that:
- The nth term consists of 'a's followed by 'b's.
- The number of 'a's in the nth term is equal to the number of 'b's in the (n-1)th term plus 1.
- The number of 'b's in the nth term is equal to the number of 'b's in the (n-1)th term plus the number of 'a's in the (n-1)th term.

### Step 2: Count the Number of 'a's
To find the total number of 'a's used from the 1st term to the 2004th term:
- The number of 'a's in each term can be represented as:
- Term 1: 1 'a'
- Term 2: 1 'a'
- Term 3: 2 'a's
- Term 4: 2 'a's
- Term 5: 3 'a's
- Term 6: 3 'a's
- ...

The pattern shows that:
- For every two terms, the number of 'a's increases by 1.

To find the total number of 'a's up to the 2004th term:
- There are 2004 terms, which means there are 1002 pairs of terms.
- Each pair contributes 1 additional 'a' to the total count.

Thus, the total number of 'a's is:
\[ \text{Total 'a's} = 1 + 1 + 2 + 2 + 3 + 3 + ... + 1002 = 2 \times (1 + 2 + 3 + ... + 1002) \]
Using the formula for the sum of the first n natural numbers:
\[ \text{Sum} = \frac{n(n + 1)}{2} \]
Substituting \( n = 1002 \):
\[ \text{Sum} = \frac{1002 \times 1003}{2} = 502503 \]
Thus, the total number of 'a's is:
\[ \text{Total 'a's} = 2 \times 502503 = 1005006 \]

### Step 3: Count the Number of 'b's
To find the total number of 'b's used from the 1st term to the 2004th term:
- The number of 'b's in each term can be represented as:
- Term 1: 0 'b's
- Term 2: 1 'b'
- Term 3: 1 'b'
- Term 4: 2 'b's
- Term 5: 2 'b's
- Term 6: 3 'b's
- ...

From the pattern:
- For every two terms, the number of 'b's increases by 1.

Thus, the total number of 'b's is:
\[ \text{Total 'b's} = 0 + 1 + 1 + 2 + 2 + 3 + 3 + ... + 1001 \]
This can be calculated similarly:
- The total number of 'b's can be calculated as:
\[ \text{Total 'b's} = 1 + 2 + 3 + ... + 1001 = \frac{1001 \times 1002}{2} = 501501 \]

### Step 4: Final Count
Now, we can summarize:
- Total 'a's = 1005006
- Total 'b's = 501501

Thus, the total number of times 'a's and 'b's are used from the 1st to the 2004th term is:
\[ \text{Total} = 1005006 + 501501 = 1506507 \]

### Final Answer
The total number of times 'a's and 'b's are used from the 1st to the 2004th term is **1506507**.