Question

Write an expression for the \(n\) th term of the sequence. (There is more than one correct answer.) \(1, \frac{1}{2}, \frac{1}{6}, \frac{1}{24}, \frac{1}{120}, \ldots\)

Short answer

The \(n^{th}\) term of the sequence can be expressed as \(\frac{1}{n!}\).

Step-by-step solution

Identifying the pattern

Analyzing the sequence, it seems that each term is of the form \(\frac{a}{n!}\). This formulates the hypothesis that the sequence is related to factorials. For instance, \(1=1!, \frac{1}{2} = \frac{1}{2!}, \frac{1}{6} = \frac{1}{3!}, \frac{1}{24} = \frac{1}{4!}, \frac{1}{120} = \frac{1}{5!}\). The relationship seems to be \(an = \frac{1}{n!}\).

Formulating the nth term

We can express the \(n^{th}\) term as \(a_n = \frac{1}{n!}\). The term \(a_n\) represents the \(n^{th}\) term in the sequence. Here, '!' indicates a factorial, meaning the product of all positive integers up to that number. For instance, \(3! = 3*2*1 = 6\). This formula gives a method to determine any term in the sequence without knowing its previous terms.

Verifying the formula

Just to be sure of the formula, let's validate this by checking with an item in the sequence, say the fifth term. Substituting \(n = 5\) in \(a_n = \frac{1}{n!}\), we get \(a_5 = \frac{1}{5!} = \frac{1}{120}\). This matches the fifth term in the given sequence. Hence, the formula holds true for the sequence.