Question
Which sequence has the given subscript notation a1=6
; an=an−1+2
when defined recursively?
; an=an−1+2
when defined recursively?
Answers
GPT-4o mini
The sequence defined by the given recursive formula can be analyzed step-by-step. The notation states:
- \( a_1 = 6 \)
- \( a_n = a_{n-1} + 2 \) for \( n > 1 \)
This means that each term in the sequence is obtained by adding 2 to the previous term.
Let's calculate the first several terms:
1. **First term** (given):
\( a_1 = 6 \)
2. **Second term**:
\( a_2 = a_1 + 2 = 6 + 2 = 8 \)
3. **Third term**:
\( a_3 = a_2 + 2 = 8 + 2 = 10 \)
4. **Fourth term**:
\( a_4 = a_3 + 2 = 10 + 2 = 12 \)
5. **Fifth term**:
\( a_5 = a_4 + 2 = 12 + 2 = 14 \)
6. **Sixth term**:
\( a_6 = a_5 + 2 = 14 + 2 = 16 \)
From this calculation, we see that the first few terms of the sequence are:
- \( a_1 = 6 \)
- \( a_2 = 8 \)
- \( a_3 = 10 \)
- \( a_4 = 12 \)
- \( a_5 = 14 \)
- \( a_6 = 16 \)
The pattern emerging here is that the sequence is an arithmetic sequence starting at 6, with a common difference of 2.
We can express the nth term of the sequence in formulaic terms as follows:
\[
a_n = 6 + 2(n - 1)
\]
Simplifying this, we get:
\[
a_n = 6 + 2n - 2 = 2n + 4
\]
Thus, the explicit form of the sequence is:
\[
a_n = 2n + 4
\]
In conclusion, the sequence defined by the recursive relation \( a_1 = 6 \) and \( a_n = a_{n-1} + 2 \) produces the terms \( 6, 8, 10, 12, 14, 16, \ldots \) which can be expressed by the formula \( a_n = 2n + 4 \).
- \( a_1 = 6 \)
- \( a_n = a_{n-1} + 2 \) for \( n > 1 \)
This means that each term in the sequence is obtained by adding 2 to the previous term.
Let's calculate the first several terms:
1. **First term** (given):
\( a_1 = 6 \)
2. **Second term**:
\( a_2 = a_1 + 2 = 6 + 2 = 8 \)
3. **Third term**:
\( a_3 = a_2 + 2 = 8 + 2 = 10 \)
4. **Fourth term**:
\( a_4 = a_3 + 2 = 10 + 2 = 12 \)
5. **Fifth term**:
\( a_5 = a_4 + 2 = 12 + 2 = 14 \)
6. **Sixth term**:
\( a_6 = a_5 + 2 = 14 + 2 = 16 \)
From this calculation, we see that the first few terms of the sequence are:
- \( a_1 = 6 \)
- \( a_2 = 8 \)
- \( a_3 = 10 \)
- \( a_4 = 12 \)
- \( a_5 = 14 \)
- \( a_6 = 16 \)
The pattern emerging here is that the sequence is an arithmetic sequence starting at 6, with a common difference of 2.
We can express the nth term of the sequence in formulaic terms as follows:
\[
a_n = 6 + 2(n - 1)
\]
Simplifying this, we get:
\[
a_n = 6 + 2n - 2 = 2n + 4
\]
Thus, the explicit form of the sequence is:
\[
a_n = 2n + 4
\]
In conclusion, the sequence defined by the recursive relation \( a_1 = 6 \) and \( a_n = a_{n-1} + 2 \) produces the terms \( 6, 8, 10, 12, 14, 16, \ldots \) which can be expressed by the formula \( a_n = 2n + 4 \).