To find the equation for the given sequence, we need to identify how the terms relate to their positions (n).
The sequence is: 1, -4, 16, ...
Let's analyze the positions and term values:
- For n = 1: a1 = 1
- For n = 2: a2 = -4
- For n = 3: a3 = 16
Next, we look for a pattern in these terms. Let's write down the values corresponding to each term:
- 1 = 1^2
- -4 = -2^2
- 16 = 4^2
The terms seem to alternate between squares of odd integers and the negatives of squares of even integers:
- For n = 1 (odd): \( (1)^2 = 1 \)
- For n = 2 (even): \( -(2)^2 = -4 \)
- For n = 3 (odd): \( (4)^2 = 16 \)
We can express the pattern mathematically:
For odd n: \[ a_n = ((n-1) + 1)^2 = n^2 \quad \text{if } n \text{ is odd} \]
For even n: \[ a_n = -(n^2) \quad \text{if } n \text{ is even} \]
A complete way to capture this in a single formula using an alternating sign based on whether n is even or odd can be described as: \[ a_n = (-1)^{n+1} \cdot ((-1)^{n-1})^2 \cdot n^2 \]
However, let's simplify it directly based on the established pattern:
Thus, we can represent the sequence as: \[ a_n = (-1)^{n} \cdot ((n-1)^2 \text{ (even nth position)}\text{ or } n^2 \text{ (odd nth position)}) \]
But for simplicity, let's represent it into a single format: If we align them in terms of n: \[ a_n = (-1)^{n+1} \cdot (n^2) \]
Thus, a valid general representation for \( a_n \) accounting for the alternating sign can be:
\[ a_n = (-1)^{n+1} \cdot \left(\frac{(n^2 + (n-1)^2)}{2}\right) \]
Thus, ultimately our answer would be: \[ a_n = (-1)^{n+1}(n^2) \] as n equals to 1 for the first term and so forth continues to operate for the subsequent terms in sequence.
3 answers