Problem 51 - College Algebra

Question

For the following exercises, graph the first five terms of the indicated sequence $$ a_{n}=\frac{(n+1) !}{(n-1) !} $$

Step-by-Step Solution

Answer

Plot the terms: \( (1, 2), (2, 6), (3, 12), (4, 20), (5, 30) \).

Step 1: Identify the Sequence Formula

The sequence is given by the formula \( a_n = \frac{(n+1)!}{(n-1)!} \). Our task is to find and graph the first five terms of this sequence.

Step 2: Calculate the First Term

To calculate the first term, \( a_1 \), substitute \( n = 1 \) into the sequence formula: \[ a_1 = \frac{(1+1)!}{(1-1)!} = \frac{2!}{0!} = \frac{2}{1} = 2. \]

Step 3: Calculate the Second Term

Substitute \( n = 2 \) into the sequence formula to find \( a_2 \): \[ a_2 = \frac{(2+1)!}{(2-1)!} = \frac{3!}{1!} = \frac{6}{1} = 6. \]

Step 4: Calculate the Third Term

Substitute \( n = 3 \) into the sequence formula to find \( a_3 \): \[ a_3 = \frac{(3+1)!}{(3-1)!} = \frac{4!}{2!} = \frac{24}{2} = 12. \]

Step 5: Calculate the Fourth Term

Substitute \( n = 4 \) into the sequence formula to find \( a_4 \): \[ a_4 = \frac{(4+1)!}{(4-1)!} = \frac{5!}{3!} = \frac{120}{6} = 20. \]

Step 6: Calculate the Fifth Term

Substitute \( n = 5 \) into the sequence formula to find \( a_5 \): \[ a_5 = \frac{(5+1)!}{(5-1)!} = \frac{6!}{4!} = \frac{720}{24} = 30. \]

Step 7: Graph the First Five Terms

Now that we have the first five terms of the sequence: \( a_1 = 2 \), \( a_2 = 6 \), \( a_3 = 12 \), \( a_4 = 20 \), \( a_5 = 30 \). Plot these values on a graph with the sequence index on the x-axis and the term value on the y-axis.

Key Concepts

Factorials

Factorials are a fundamental concept in mathematics, particularly in algebra and sequences. It involves multiplying a series of descending natural numbers. For any positive integer \( n \), the factorial, written as \( n! \), is the product \( n \times (n-1) \times (n-2) \times \, ... \, \times 1 \). Factorials are useful in sequences because they can help define complex relationships between terms. In the sequence given by \( a_n = \frac{(n+1)!}{(n-1)!} \), the factorials \((n+1)!\) and \((n-1)!\) enable the simplification of terms. This simplification is crucial in determining each sequence term efficiently. One important property to remember is \( 0! = 1 \), which is handy in calculations like \( \frac{2!}{0!} = \frac{2}{1} = 2 \). Factorials can grow very large quickly, thus making them interesting for defining sequences that grow rapidly.

Graphing Sequences

Graphing sequences helps us visualize the pattern or growth of terms. To graph a sequence, we plot each term against its index (or position) in the sequence. For example, in the sequence \( a_n = \frac{(n+1)!}{(n-1)!} \), the first step is to calculate each term, as done in the solution steps. Each calculated term forms a point on the graph. The x-axis usually represents the index \( n \), and the y-axis represents the term value \( a_n \).

• First term: \((1, 2)\)
• Second term: \((2, 6)\)
• Third term: \((3, 12)\)
• Fourth term: \((4, 20)\)
• Fifth term: \((5, 30)\)

Plotting these on a graph can help identify trends like increases or decreases between terms. In this specific sequence, as \( n \) increases, the values of the terms also increase rapidly, confirming the factorial's influence in the sequence dynamics.

Term Calculation

Calculating terms in a sequence involves substituting specific values into the sequence's formula. For the given sequence \( a_n = \frac{(n+1)!}{(n-1)!} \), term calculation starts with identifying the appropriate \( n \) value you wish to compute. For instance, to find \( a_1 \), substitute \( n = 1 \) into the formula and simplify: \[ a_1 = \frac{(1+1)!}{(1-1)!} = \frac{2!}{0!} = 2. \] This calculation continues for each desired index:

• For \( a_2 \), substitute \( n = 2 \) to get \( 6 \).
• For \( a_3 \), substitute \( n = 3 \) to get \( 12 \).
• For \( a_4 \), substitute \( n = 4 \) to get \( 20 \).
• For \( a_5 \), substitute \( n = 5 \) to get \( 30 \).

The formula captures the sequence's structure, and term calculation helps set up the graph or understand the sequence's pathways. Each substitution follows mathematical laws and leverages factorial relationships to achieve accurate term values.