Question

write an expression for the nth term of the following sequences: (assume n=1, 2, 3, …)

1. 3, 7, 11, 15, …
2. 1, \frac{-1}{2}, \frac{1}{6}, \frac{-1}{24}, \frac{1}{120}, …
3. 2, 6, 18, 54, …

Explanation:

Problem 26: Sequence \( 3, 7, 11, 15, \dots \)

This is an arithmetic sequence. In an arithmetic sequence, the \( n \)-th term is given by \( a_n = a_1 + (n - 1)d \), where \( a_1 \) is the first term and \( d \) is the common difference.

Step 1: Identify \( a_1 \) and \( d \)

The first term \( a_1 = 3 \). The common difference \( d \) is the difference between consecutive terms. For example, \( 7 - 3 = 4 \), \( 11 - 7 = 4 \), \( 15 - 11 = 4 \), so \( d = 4 \).

Step 2: Substitute into the arithmetic sequence formula

Using \( a_n = a_1 + (n - 1)d \), substitute \( a_1 = 3 \) and \( d = 4 \):
\[
a_n = 3 + (n - 1) \times 4
\]

Step 3: Simplify the expression

\[
a_n = 3 + 4n - 4 = 4n - 1
\]

Let's analyze the pattern of signs, numerators, and denominators.

Step 1: Analyze the sign pattern

The signs alternate: positive, negative, positive, negative, positive, ... This can be represented by \( (-1)^{n + 1} \) (since for \( n = 1 \), \( (-1)^{1 + 1} = 1 \); for \( n = 2 \), \( (-1)^{2 + 1} = -1 \), etc.).

Step 2: Analyze the denominator pattern

The denominators are \( 1, 2, 6, 24, 120, \dots \). Notice that \( 1 = 1! \), \( 2 = 2! \), \( 6 = 3! \), \( 24 = 4! \), \( 120 = 5! \), so the denominator of the \( n \)-th term is \( n! \).

Step 3: Analyze the numerator pattern

The numerators are \( 1, 1, 1, 1, 1, \dots \) (since the absolute value of each term's numerator is 1).

Step 4: Combine the patterns

Putting it all together, the \( n \)-th term is:
\[
a_n = \frac{(-1)^{n + 1}}{n!}
\]

This is a geometric sequence. In a geometric sequence, the \( n \)-th term is given by \( a_n = a_1 \times r^{n - 1} \), where \( a_1 \) is the first term and \( r \) is the common ratio.

Step 1: Identify \( a_1 \) and \( r \)

The first term \( a_1 = 2 \). The common ratio \( r \) is the ratio of consecutive terms. For example, \( \frac{6}{2} = 3 \), \( \frac{18}{6} = 3 \), \( \frac{54}{18} = 3 \), so \( r = 3 \).

Step 2: Substitute into the geometric sequence formula

Using \( a_n = a_1 \times r^{n - 1} \), substitute \( a_1 = 2 \) and \( r = 3 \):
\[
a_n = 2 \times 3^{n - 1}
\]

Answer:

The \( n \)-th term is \( \boldsymbol{a_n = 4n - 1} \)

Problem 27: Sequence \( 1, \frac{-1}{2}, \frac{1}{6}, \frac{-1}{24}, \frac{1}{120}, \dots \)