Question

test.
in exercises 1 and 2, find the first four terms and the fortieth term of
the given sequence.

1. $a_n = (-1)^n \frac{n + 1}{n + 3}$ for all $n \geq 1$
2. $a_1 = -3, a_n = 2a_{n - 1}$ for all $n \geq 2$
3. the sequence $-1, 1/2, 2, 7/2, \dots$ is arithmetic. find (a) the com-

mon difference, (b) the tenth term, and (c) an explicit rule for the
$n$th term.

1. the sequence $1/2, -2, 8, -32, \dots$ is geometric. find (a) the

common ratio, (b) the seventh term, and (c) an explicit rule for
the $n$th term.

Answer:

Let's solve each problem one by one. We'll start with problem 1.

Problem 1: \( a_n = (-1)^n \frac{n + 1}{n + 3} \) for \( n \geq 1 \)

We need to find the first four terms and the fortieth term.

Step 1: Find the first term (\( n = 1 \))

Substitute \( n = 1 \) into the formula:
\( a_1 = (-1)^1 \frac{1 + 1}{1 + 3} = -1 \cdot \frac{2}{4} = -\frac{1}{2} \)

Step 2: Find the second term (\( n = 2 \))

Substitute \( n = 2 \):
\( a_2 = (-1)^2 \frac{2 + 1}{2 + 3} = 1 \cdot \frac{3}{5} = \frac{3}{5} \)

Step 3: Find the third term (\( n = 3 \))

Substitute \( n = 3 \):
\( a_3 = (-1)^3 \frac{3 + 1}{3 + 3} = -1 \cdot \frac{4}{6} = -\frac{2}{3} \)

Step 4: Find the fourth term (\( n = 4 \))

Substitute \( n = 4 \):
\( a_4 = (-1)^4 \frac{4 + 1}{4 + 3} = 1 \cdot \frac{5}{7} = \frac{5}{7} \)

Step 5: Find the fortieth term (\( n = 40 \))

Substitute \( n = 40 \):
\( a_{40} = (-1)^{40} \frac{40 + 1}{40 + 3} = 1 \cdot \frac{41}{43} = \frac{41}{43} \)

Problem 2: \( a_1 = -3 \), \( a_n = 2a_{n - 1} \) for \( n \geq 2 \)

This is a geometric sequence with first term \( a_1 = -3 \) and common ratio \( r = 2 \).

Step 1: Find the first term

\( a_1 = -3 \)

Step 2: Find the second term (\( n = 2 \))

\( a_2 = 2a_1 = 2(-3) = -6 \)

Step 3: Find the third term (\( n = 3 \))

\( a_3 = 2a_2 = 2(-6) = -12 \)

Step 4: Find the fourth term (\( n = 4 \))

\( a_4 = 2a_3 = 2(-12) = -24 \)

Step 5: Find the fortieth term

The formula for the \( n \)-th term of a geometric sequence is \( a_n = a_1 r^{n - 1} \).
Substitute \( a_1 = -3 \), \( r = 2 \), and \( n = 40 \):
\( a_{40} = -3 \cdot 2^{39} \)

Problem 3: Arithmetic sequence \( -1, \frac{1}{2}, 2, \frac{7}{2}, \dots \)

An arithmetic sequence has the form \( a_n = a_1 + (n - 1)d \), where \( d \) is the common difference.

(a) Find the common difference \( d \)

The common difference \( d \) is the difference between consecutive terms.
\( d = \frac{1}{2} - (-1) = \frac{1}{2} + 1 = \frac{3}{2} \)
Check with the next term: \( 2 - \frac{1}{2} = \frac{3}{2} \), and \( \frac{7}{2} - 2 = \frac{3}{2} \). So \( d = \frac{3}{2} \).

(b) Find the tenth term

Use the arithmetic sequence formula: \( a_n = a_1 + (n - 1)d \)
\( a_1 = -1 \), \( d = \frac{3}{2} \), \( n = 10 \)
\( a_{10} = -1 + (10 - 1)\frac{3}{2} = -1 + 9 \cdot \frac{3}{2} = -1 + \frac{27}{2} = \frac{-2 + 27}{2} = \frac{25}{2} \)

(c) Find an explicit rule for the \( n \)-th term

Using \( a_n = a_1 + (n - 1)d \):
\( a_n = -1 + (n - 1)\frac{3}{2} = -1 + \frac{3}{2}n - \frac{3}{2} = \frac{3}{2}n - \frac{5}{2} \)

Problem 4: Geometric sequence \( \frac{1}{2}, -2, 8, -32, \dots \)

A geometric sequence has the form \( a_n = a_1 r^{n - 1} \), where \( r \) is the common ratio.

(a) Find the common ratio \( r \)

The common ratio \( r \) is the ratio of consecutive terms.
\( r = \frac{-2}{\frac{1}{2}} = -2 \cdot 2 = -4 \)
Check with the next term: \( \frac{8}{-2} = -4 \), and \( \frac{-32}{8} = -4 \). So \( r = -4 \).

(b) Find the seventh term

First, find \( a_1 = \frac{1}{2} \), \( r = -4 \), \( n = 7 \)
Using the geometric sequence formula: \( a_n = a_1 r^{n - 1} \)
\( a_7 = \frac{1}{2} \cdot (-4)^{6} = \frac{1}{2} \cdot 4096 = 2048 \)

(c) Find an explicit rule for the \( n \)-th term

Using \( a_n = a_1 r^{n - 1} \):
\( a_n = \frac{1}{2} \cdot (-4)^{n - 1} \)
We can also simplify this:
\( (-4)^{n - 1} = (-1)^{n - 1} \cdot 4^{n - 1} \)
So \( a_n = \frac{1}{2} \cdot (-1)^{n - 1} \cdot 4^{n - 1} = (-1)^{n - 1} \cdot 2^{2(n - 1) - 1} = (-1)^{n - 1} \cdot 2^{2n - 3} \)
Or, alternatively, \( a_n = \frac{1}{2}(-4)^{n - 1} \) is also correct.