Question
Write a recursive formula for a, start subscript, n, end subscripta
n
, the n, start superscript, th, end superscriptn
th
term of the sequence 50, comma, minus, 10, comma, 2, comma, dot dot dot50,−10,2,....
Answer
Attempt 1 out of 2
a, start subscript, 1, end subscript, equalsa
1
=
a, start subscript, n, end subscript, equalsa
n
=
n
, the n, start superscript, th, end superscriptn
th
term of the sequence 50, comma, minus, 10, comma, 2, comma, dot dot dot50,−10,2,....
Answer
Attempt 1 out of 2
a, start subscript, 1, end subscript, equalsa
1
=
a, start subscript, n, end subscript, equalsa
n
=
Answers
Answered by
GPT-4o mini
To find a recursive formula for the sequence \( a_n \) given the terms \( 50, -10, 2, \ldots \), we first need to establish the relationship between consecutive terms.
Looking at the sequence, the pattern appears to be:
- \( a_1 = 50 \)
- \( a_2 = -10 \)
- \( a_3 = 2 \)
Now let's calculate the differences between consecutive terms:
- \( a_2 - a_1 = -10 - 50 = -60 \)
- \( a_3 - a_2 = 2 - (-10) = 2 + 10 = 12 \)
Next, let's see if we can identify a possible recurrence.
From the terms provided, we can observe:
- We subtract a value from 50 to get to -10 (which is 60), but rather than focusing on plain differences, let's look at how we can relate \( a_n \) to its previous terms.
One way to express this recursively is:
1. Identify a consistent transformation of one term to the next.
2. Notice that it appears to alternate in sign and the amount added/subtracted seems to vary.
Here's a potential recursive relationship based on what we've seen:
- \( a_1 = 50 \)
- For \( n \geq 2 \):
\[
a_n =
\begin{cases}
a_{n-1} - 60 & \text{if } n \text{ is even}\\
a_{n-1} + 12 & \text{if } n \text{ is odd}
\end{cases}
\]
However, upon further inspection, it seems a more consistent and simple approach might just be to define it directly in terms of the previous term and incorporate adjustments by a constant.
We can formalize it as:
\[
a_n = a_{n-1} + d \text{ for some constant } d
\]
To summarize:
\[
a_1 = 50
\]
\[
a_n = a_{n-1} - 60 + 12 \text{ (alternating decrease/increase)}
\]
or we could choose simply:
\[
a_n = a_{n-1} + (-60 \text{ or } 12 \text{ depending on odd/even terms})
\]
For more simplicity, it could be generalized, but the important part is recognizing:
\[
a_n = a_{n-1} - 60 + (12 if n > 2, else just equall the sign flip after
a_{n-1} )
\]
is a way to express recursive relationship.
Ultimately:
\[
\text{Answer:} \\
a_1 = 50 \\
a_n = a_{n-1} + d_n \text{ where } d_n \text{ is the change as determined by odd/even}
\]
is a correct summary.
Looking at the sequence, the pattern appears to be:
- \( a_1 = 50 \)
- \( a_2 = -10 \)
- \( a_3 = 2 \)
Now let's calculate the differences between consecutive terms:
- \( a_2 - a_1 = -10 - 50 = -60 \)
- \( a_3 - a_2 = 2 - (-10) = 2 + 10 = 12 \)
Next, let's see if we can identify a possible recurrence.
From the terms provided, we can observe:
- We subtract a value from 50 to get to -10 (which is 60), but rather than focusing on plain differences, let's look at how we can relate \( a_n \) to its previous terms.
One way to express this recursively is:
1. Identify a consistent transformation of one term to the next.
2. Notice that it appears to alternate in sign and the amount added/subtracted seems to vary.
Here's a potential recursive relationship based on what we've seen:
- \( a_1 = 50 \)
- For \( n \geq 2 \):
\[
a_n =
\begin{cases}
a_{n-1} - 60 & \text{if } n \text{ is even}\\
a_{n-1} + 12 & \text{if } n \text{ is odd}
\end{cases}
\]
However, upon further inspection, it seems a more consistent and simple approach might just be to define it directly in terms of the previous term and incorporate adjustments by a constant.
We can formalize it as:
\[
a_n = a_{n-1} + d \text{ for some constant } d
\]
To summarize:
\[
a_1 = 50
\]
\[
a_n = a_{n-1} - 60 + 12 \text{ (alternating decrease/increase)}
\]
or we could choose simply:
\[
a_n = a_{n-1} + (-60 \text{ or } 12 \text{ depending on odd/even terms})
\]
For more simplicity, it could be generalized, but the important part is recognizing:
\[
a_n = a_{n-1} - 60 + (12 if n > 2, else just equall the sign flip after
a_{n-1} )
\]
is a way to express recursive relationship.
Ultimately:
\[
\text{Answer:} \\
a_1 = 50 \\
a_n = a_{n-1} + d_n \text{ where } d_n \text{ is the change as determined by odd/even}
\]
is a correct summary.