Problem 20 - College Algebra

Question

The general term of a sequence is given and involves a factorial. Write the first four terms of each sequence. $$a_{n}=\frac{(n+1) !}{n^{2}}$$

Step-by-Step Solution

Answer

The first four terms of the sequence are 2, 1.5, 2.67, and 7.5

Step 1: Understanding Factorial

The factorial of a non-negative integer \(n\), denoted by \(n!\), is the product of all positive integers less than or equal to \(n\). For example, 5! = 5×4×3×2×1 = 120.

Step 2: Calculate First Term

Substitute 1 in place of \(n\) in the equation \(a_{n}=\frac{(n+1)!}{n^{2}}\), we get \(a_{1}=\frac{(1+1)!}{1^{2}} = \frac{2!}{1} = \frac{2}{1} = 2\)

Step 3: Calculate Second Term

Substitute 2 in place of \(n\) in the equation \(a_{n}=\frac{(n+1)!}{n^{2}}\), we get \(a_{2}=\frac{(2+1)!}{2^{2}} = \frac{3!}{4} = \frac{6}{4} = 1.5\)

Step 4: Calculate Third Term

Substitute 3 in place of \(n\) in the equation \(a_{n}=\frac{(n+1)!}{n^{2}}\), we get \(a_{3}=\frac{(3+1)!}{3^{2}} = \frac{4!}{9} = \frac{24}{9} = 2.67\)

Step 5: Calculate Fourth Term

Substitute 4 in place of \(n\) in the equation \(a_{n}=\frac{(n+1)!}{n^{2}}\), we get \(a_{4}=\frac{(4+1)!}{4^{2}} = \frac{5!}{16} = \frac{120}{16} = 7.5\)

Step 6: Formulate the Sequence

Putting the calculated terms together, the first four terms of the sequence are: 2, 1.5, 2.67, 7.5