Determine the first four terms of the sequence in which the nth term is a subscript n baseline equals Start Fraction left-parenthesis n plus 1 right-parenthesis exclamation-mark over left-parenthesis n plus 2 right-parenthesis exclamation-mark End Fraction.

Question

(1 point)
Responses

one-half comma one-third comma one-fourth comma Start Fraction 1 over 5 End Fraction
Image with alt text: one-half comma one-third comma one-fourth comma Start Fraction 1 over 5 End Fraction

Start Fraction 1 over 3 End Fraction comma Start Fraction 1 over 4 End Fraction comma Start Fraction 1 over 5 End Fraction comma Start Fraction 1 over 6 End Fraction
Image with alt text: Start Fraction 1 over 3 End Fraction comma Start Fraction 1 over 4 End Fraction comma Start Fraction 1 over 5 End Fraction comma Start Fraction 1 over 6 End Fraction

one-half comma one-half comma one-half comma one-half
Image with alt text: one-half comma one-half comma one-half comma one-half

Start Fraction 2 over 3 End Fraction comma Start Fraction 3 over 4 End Fraction comma Start Fraction 4 over 5 End Fraction comma Start Fraction 5 over 6 End Fraction

Answer 1

To determine the first four terms of the sequence given by the formula:

\[ a_n = \frac{(n + 1)!}{(n + 2)!} \]

we need to compute \(a_n\) for \(n = 1, 2, 3, 4\).

First, let's simplify \(a_n\):

\[ a_n = \frac{(n + 1)!}{(n + 2)(n + 1)!} = \frac{1}{n + 2} \]

Now we can find the first four terms:

1. For \(n = 1\): \[ a_1 = \frac{1}{1 + 2} = \frac{1}{3} \]

2. For \(n = 2\): \[ a_2 = \frac{1}{2 + 2} = \frac{1}{4} \]

3. For \(n = 3\): \[ a_3 = \frac{1}{3 + 2} = \frac{1}{5} \]

4. For \(n = 4\): \[ a_4 = \frac{1}{4 + 2} = \frac{1}{6} \]

Hence, the first four terms of the sequence are:

\[ \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6} \]

Thus, the correct response is:

\(\frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}\)