Write the recursive formula for the following sequence {0.25, 0.75, 2.25, 6.75 …}
The Correct answer and Explanation is :
The correct answer is: a_1 = 0.25
To find the recursive formula for the sequence ({0.25, 0.75, 2.25, 6.75, \ldots}), we first need to identify the pattern in the sequence and how each term relates to its predecessor.
Identifying the Pattern
Let’s denote the (n)-th term of the sequence as (a_n):
- (a_1 = 0.25)
- (a_2 = 0.75)
- (a_3 = 2.25)
- (a_4 = 6.75)
Next, we will observe the ratios of successive terms:
- (\frac{a_2}{a_1} = \frac{0.75}{0.25} = 3)
- (\frac{a_3}{a_2} = \frac{2.25}{0.75} = 3)
- (\frac{a_4}{a_3} = \frac{6.75}{2.25} = 3)
It appears that each term is multiplied by 3 as we move from one term to the next. However, simply multiplying by 3 doesn’t give us a straightforward recursive relationship. Instead, we can examine the differences between the terms:
- (a_2 – a_1 = 0.75 – 0.25 = 0.50)
- (a_3 – a_2 = 2.25 – 0.75 = 1.50)
- (a_4 – a_3 = 6.75 – 2.25 = 4.50)
The differences themselves appear to follow a pattern:
- The differences are (0.50, 1.50, 4.50), which can be rewritten as (0.5 \cdot 1, 0.5 \cdot 3, 0.5 \cdot 9).
The pattern in the differences seems to indicate that the change from one term to the next involves multiplying the previous term by 3 and then subtracting a value that is also increasing as powers of 3.
Recursive Formula
From our observations, we can formulate the recursive equation as follows:
[
a_n = 3a_{n-1} – 0.5 \cdot 3^{n-2} \quad \text{for } n \geq 2
]
With the initial term given by:
[
a_1 = 0.25
]
Explanation
In this recursive formula, each term (a_n) is generated by taking three times the previous term (a_{n-1}) and adjusting it by subtracting a scaled value that reflects the rapid growth of the sequence. The presence of (3^{n-2}) indicates that the influence of the previous terms grows exponentially as (n) increases, creating a sequence that accelerates in growth. This understanding helps recognize that while multiplication drives the growth, the subtractive term accounts for the deviations seen in the sequence, providing a complete representation of how each term is generated based on its predecessor.