Write the set `B={a_(n):n in N,a_(n+2)=a_(n+1)+a_(n)anda_(1)=a_(2)=1}` in tabular form.

Answer

To write the set \( B = \{ a_n : n \in \mathbb{N}, a_{n+2} = a_{n+1} + a_n, a_1 = a_2 = 1 \} \) in tabular form, we need to calculate the values of \( a_n \) for the first few natural numbers \( n \).

### Step-by-Step Solution:

1. **Identify the initial conditions**:
- We know that \( a_1 = 1 \) and \( a_2 = 1 \).

2. **Calculate the next terms using the recurrence relation**:
- The recurrence relation is given by \( a_{n+2} = a_{n+1} + a_n \).
- We can calculate the next terms as follows:
- \( a_3 = a_2 + a_1 = 1 + 1 = 2 \)
- \( a_4 = a_3 + a_2 = 2 + 1 = 3 \)
- \( a_5 = a_4 + a_3 = 3 + 2 = 5 \)
- \( a_6 = a_5 + a_4 = 5 + 3 = 8 \)

3. **List the values in tabular form**:
- Now we can list the values of \( a_n \) for \( n = 1, 2, 3, 4, 5, 6 \):
\[ \begin{array}{|c|c|} \hline n & a_n \\ \hline 1 & 1 \\ 2 & 1 \\ 3 & 2 \\ 4 & 3 \\ 5 & 5 \\ 6 & 8 \\ \hline \end{array} \]

4. **Continue the pattern if needed**:
- If we want to extend the table further, we can continue calculating:
- \( a_7 = a_6 + a_5 = 8 + 5 = 13 \)
- \( a_8 = a_7 + a_6 = 13 + 8 = 21 \)

5. **Final tabular form**:
- The extended table would look like this:
\[ \begin{array}{|c|c|} \hline n & a_n \\ \hline 1 & 1 \\ 2 & 1 \\ 3 & 2 \\ 4 & 3 \\ 5 & 5 \\ 6 & 8 \\ 7 & 13 \\ 8 & 21 \\ \hline \end{array} \]