Question

Write the first five terms of the sequence. \(a_{n}=\left(-\frac{1}{2}\right)^{n}\)

Short answer

The first five terms of the sequence are: -1/2, 1/4, -1/8, 1/16, -1/32

Step-by-step solution

Solve for n=1

Substitute n=1 into the formula : \(a_{1}=\left(-\frac{1}{2}\right)^{1} = -\frac{1}{2}\)

Solve for n=2

Substitute n=2 into the formula: \(a_{2}=\left(-\frac{1}{2}\right)^{2} = \frac{1}{4}\)

Solve for n=3

Substitute n=3 into the formula: \(a_{3}=\left(-\frac{1}{2}\right)^{3} = -\frac{1}{8}\)

Solve for n=4

Substitute n=4 into the formula: \(a_{4}=\left(-\frac{1}{2}\right)^{4} = \frac{1}{16}\)

Solve for n=5

Substitute n=5 into the formula: \(a_{5}=\left(-\frac{1}{2}\right)^{5} = -\frac{1}{32}\)

Key concepts

Arithmetic Sequence

An arithmetic sequence is a series of numbers in which each term after the first is obtained by adding a constant difference to the previous term. This constant is called the 'common difference'. For example, in the sequence 2, 5, 8, 11, ..., the common difference is 3 because each term is the previous term plus 3.

Let's express it formally. If the first term of an arithmetic sequence is denoted by \( a_1 \), the second by \( a_2 \), and so on, the nth term \( a_n \) can be calculated using the formula:
\[ a_n = a_1 + (n-1)d \]
where \( d \) is the common difference. It's important to notice that in an arithmetic sequence, the terms increase or decrease linearly. This means for each step forward in the sequence, the value changes by adding or subtracting the same amount, the common difference.

However, the sequence given in the exercise, \(a_{n}=\big(-\frac{1}{2}\big)^{n}\), does not have a common difference between terms as each term is a result of exponentiation, not a linear change, hence it is not an arithmetic sequence.

Geometric Sequence

In contrast, a geometric sequence is defined by each term being a multiple of the previous term. This constant factor is known as the 'common ratio'. For a simple case, consider the sequence 3, 6, 12, 24, ..., where each term is multiplied by 2 to get the next term.

The nth term of a geometric sequence with the first term \( a_1 \) and a common ratio \( r \) is given as:
\[ a_n = a_1 \times r^{(n-1)} \]
This formula tells you that to find any term in a geometric series, you take the first term and multiply it by the common ratio raised to the power of \( n-1 \), where \( n \) is the position of the term in the sequence. Geometric sequences show exponential growth or decay. If the ratio is greater than 1, the sequence grows, and if the ratio is between 0 and 1, the sequence decays.

Looking at the original exercise, the given sequence \(a_{n}=\big(-\frac{1}{2}\big)^{n}\) involves raising a number to the power of \( n \), which indicates we are dealing with a geometric sequence where the common ratio is \(-\frac{1}{2}\).

Exponentiation

Exponentiation is a mathematical operation involving two numbers, the base \( b \) and the exponent \( n \). It is written as \(b^n\) and pronounced as 'b raised to the power of n'. The exponent signifies how many times the base is multiplied by itself. For example, \(3^2\) (read as 'three squared') is 3 multiplied by itself: \( 3 \times 3 = 9 \).

Particularly noteworthy points include:

• Negative exponents represent division, as in \( b^{-n} = \frac{1}{b^n} \).
• When a base is raised to a fractional exponent, it represents roots, such as \( b^{1/n} = \root{n}\b \).
• Exponents that are not whole numbers can represent more complex operations like irrational roots.

Exponentiation is crucial in geometric sequences. The exercise illustrates this with \(a_{n}=\big(-\frac{1}{2}\big)^{n}\): each term is found by raising the common ratio \(-\frac{1}{2}\) to the power of \( n \). The alternation of sign from positive to negative is due to using a negative base with successive powers, a characteristic pattern in sequences involving exponentiation with negative bases.