Let's analyze the sequence:
The terms are:
- \( a_1 = -11 \)
- \( a_2 = \frac{22}{3} \)
- \( a_3 = -\frac{44}{9} \)
To write a formula for the \( n \)-th term \( a_n \), we first look at the numerators and denominators separately.
Step 1: Numerators
The numerators of the terms are:
- \( -11 \) (which is \( -11 = -11 \times 1 \))
- \( 22 \) (which is \( 22 = 11 \times 2 \))
- \( -44 \) (which is \( -44 = -11 \times 4 \))
It seems the numerators alternate in sign and their absolute values are \( 11, 22, 44 \):
- The pattern in the absolute values: \( | -11 | = 11 \), \( | 22 | = 22 \), \( | -44 | = 44 \).
- The absolute values correspond to \( 11 \times 2^{n-1} \) since:
- \( a_1: 11 \times 2^{0} = 11 \)
- \( a_2: 11 \times 2^{1} = 22 \)
- \( a_3: 11 \times 2^{2} = 44 \)
The sign alternates starting with negative. We can incorporate this into our equation by alternating signs using \( (-1)^n \):
- \( (-1)^n \cdot 11 \cdot 2^{n-1} \)
Step 2: Denominators
Next, consider the denominators:
- For \( n = 1 \), no denominator (which is \( 1 \))
- For \( n = 2 \), \( 3 \)
- For \( n = 3 \), \( 9 \)
It looks like the denominators are powers of 3:
- For \( n = 1 \), \( 3^{0} = 1 \)
- For \( n = 2 \), \( 3^{1} = 3 \)
- For \( n = 3 \), \( 3^{2} = 9 \)
Thus, the denominator can be expressed as \( 3^{n-1} \).
Putting it all together
So, the general term \( a_n \) can be expressed as: \[ a_n = \frac{(-1)^n \cdot 11 \cdot 2^{n-1}}{3^{n-1}} \]
You can also express it in a more concise way: \[ a_n = (-1)^n \cdot \frac{11 \cdot 2^{n-1}}{3^{n-1}} \]
This gives us the equation for the \( n \)-th term in the sequence.
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