What if you’re given a sequence and need to figure out the explicit formula for it?
In this case, you can work backward to discover the formula using the first few terms. The key is to look for patterns in the sequence. These patterns might involve alternating terms, or there could be specific formulas for the numerators, denominators, exponents, or bases. Once you spot the pattern, you can use it to create the explicit formula for the entire sequence.
Write an explicit formula for the [latex]n\text{th}[/latex] term of each sequence.
[latex]\begin{align} & \left\{-\dfrac{2}{25},-\dfrac{2}{125},-\dfrac{2}{625},-\dfrac{2}{3\text{,}125},-\dfrac{2}{15\text{,}625},\dots \right\}&& \text{The numerator is 2.} \\[1mm] & \left\{-\dfrac{2}{5^2},-\dfrac{2}{5^3},- \dfrac{2}{5^4},-\dfrac{2}{5^5},-\dfrac{2}{5^6},\dots,-\dfrac{2}{5^n},\dots \right\} && \text{The denominators are increasing powers of 5.}\end{align}[/latex]
So we know that the fraction is negative, the numerator is [latex]2[/latex], and the denominator can be represented by [latex]{5}^{n+1}[/latex].
[latex]{a}_{n}=-\dfrac{2}{{5}^{n+1}}[/latex]
The terms are powers of [latex]e[/latex]. For [latex]n=1[/latex], the first term is [latex]{e}^{4}[/latex] so the exponent must be [latex]n+3[/latex].
[latex]{a}_{n}={e}^{n+3}[/latex]
Alternating Sequences
Sometimes sequences have terms that are alternate. In fact, the terms may actually alternate in sign. The steps to finding terms of the sequence are the same as if the signs did not alternate.
Let’s take a look at the following sequence.
[latex]\left\{2,-4,6,-8\right\}[/latex]
Notice the first term is greater than the second term, the second term is less than the third term, and the third term is greater than the fourth term. This trend continues forever. Do not rearrange the terms in numerical order to interpret the sequence.
The sign of the term in the explicit formula is given by the [latex]{\left(-1\right)}^{n}[/latex] if the first term is negative and [latex]{\left(-1\right)}^{n-1}[/latex] if the first term is positive.Write the first five terms of the sequence.
The first five terms are [latex]\left\{-\dfrac{1}{2},\dfrac{4}{3},-\dfrac{9}{4},\dfrac{16}{5},-\dfrac{25}{6}\right\}[/latex].
Analysis of the Solution
The graph of this function looks different from the ones we have seen previously in this section because the terms of the sequence alternate between positive and negative values.
Write an explicit formula for the [latex]n\text{th}[/latex] term of each sequence.
The terms alternate between positive and negative. We can use [latex]{\left(-1\right)}^{n}[/latex] to make the terms alternate. The numerator can be represented by [latex]n+1[/latex].
The denominator can be represented by [latex]2n+9[/latex].
