Problem 58 - Precalculus with Limits

Question

In Exercises \(47-62,\) write an expression for the apparent \(n\) th term of the sequence. (Assume that \(n\) begins with \(1 . )\) $$ 1,2, \frac{2^{2}}{2}, \frac{2^{3}}{6}, \frac{2^{4}}{24}, \frac{2^{5}}{120}, \ldots $$

Step-by-Step Solution

Answer

The \(n\)th term of the sequence is given by the expression \( \frac{2^n}{n!} \).

Step 1: Observation of Pattern

The first term is 1 which can be rewritten as \( \frac{2^0}{0!} \), since both \(2^0\) and \(0!\) are equal to 1. The second term is 2, which equals \( \frac{2^1}{1!} \). The third term is \( \frac{2^2}{2} \) = \( \frac{2^2}{2!} \). As we continue this for other terms, we can see a clear pattern: the \(n\)th term of the sequence appears to be \( \frac{2^n}{n!} \).

Step 2: Formulate the \(n\)th Term

Based on our observation, we can formulate the \(n\)th term of the sequence as \( \frac{2^n}{n!} \). Where \(n\) is the \(n\)th term.

Step 3: Verification through Substitution

To make sure this expression is correct, substitute different values of \(n\) in the equation and compare it with the corresponding terms of the sequence. For example, if we substitute \(n = 4\) in the equation, we get \( \frac{2^4}{4!} = \frac{16}{24} = \frac{2^4}{24} \), which corresponds to the 4th term of the sequence. Carry out this check for different values of \(n\). If it corresponds to the terms in the sequence, it can be assumed that the expression for the \(n\)th term is correct.