The first two terms of a sequence are 0 and 1, The `n ^(th)` terms `T _(n) = 2 T _(n-1) - T _(n-2) , n ge 3.` For example the third terms `T _(3) = 2 T_(2) - T_(1) = 2 -0=2,` The sum of the first 2006 terms of this sequence is :

Answer

To find the sum of the first 2006 terms of the sequence defined by \( T_n = 2T_{n-1} - T_{n-2} \) with initial terms \( T_1 = 0 \) and \( T_2 = 1 \), we will first calculate the terms of the sequence and then find the sum.

### Step-by-Step Solution:

1. **Identify the first two terms:**
\[ T_1 = 0, \quad T_2 = 1 \]

2. **Calculate the third term:**
\[ T_3 = 2T_2 - T_1 = 2 \times 1 - 0 = 2 \]

3. **Calculate the fourth term:**
\[ T_4 = 2T_3 - T_2 = 2 \times 2 - 1 = 4 - 1 = 3 \]

4. **Calculate the fifth term:**
\[ T_5 = 2T_4 - T_3 = 2 \times 3 - 2 = 6 - 2 = 4 \]

5. **Calculate the sixth term:**
\[ T_6 = 2T_5 - T_4 = 2 \times 4 - 3 = 8 - 3 = 5 \]

6. **Continue calculating the next terms to establish a pattern:**
\[ T_7 = 2T_6 - T_5 = 2 \times 5 - 4 = 10 - 4 = 6 \]
\[ T_8 = 2T_7 - T_6 = 2 \times 6 - 5 = 12 - 5 = 7 \]
\[ T_9 = 2T_8 - T_7 = 2 \times 7 - 6 = 14 - 6 = 8 \]

7. **Observe the pattern:**
The terms appear to be increasing sequentially:
\[ T_1 = 0, T_2 = 1, T_3 = 2, T_4 = 3, T_5 = 4, T_6 = 5, T_7 = 6, T_8 = 7, T_9 = 8, \ldots \]
This suggests that \( T_n = n - 1 \) for \( n \geq 1 \).

8. **Verify the formula:**
For \( n \geq 3 \):
\[ T_n = n - 1 \]
This holds true based on the calculations above.

9. **Calculate the sum of the first 2006 terms:**
The first term is \( T_1 = 0 \) and the last term is \( T_{2006} = 2005 \). Thus, we need to sum the integers from 1 to 2005:
\[ S = 1 + 2 + 3 + \ldots + 2005 \]

10. **Use the formula for the sum of the first \( n \) natural numbers:**
\[ S = \frac{n(n + 1)}{2} \]
Here, \( n = 2005 \):
\[ S = \frac{2005 \times (2005 + 1)}{2} = \frac{2005 \times 2006}{2} \]

11. **Final calculation:**
\[ S = \frac{2005 \times 2006}{2} \]

### Final Answer:
The sum of the first 2006 terms of the sequence is:
\[ \frac{2005 \times 2006}{2} \]