TBIL-Cal Sequence Formulas (SQ1)

As seen in the previous activity, having too few terms may prevent us from finding a unique way to continue creating a sequence of numbers. In fact, we need sufficiently many terms to uniquely continue a sequence of numbers (and how many terms is sufficient depends on which sequence of numbers you are trying to generate). Sometimes, we do not want to write out all of the terms needed to allow for this. Therefore, we will want to find short-hand notation that allows us to do so.

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Definition 8.1.4.

A sequence is a list of real numbers. Let \(a_n\) denote the \(n\)th term in a sequence. We will use the notation \(\displaystyle \{a_n\}_{n=1}^\infty=a_1, a_2, \ldots, a_n, \ldots\text{.}\) A general formula that indicates how to explicitly find the \(n\)-th term of a sequence is the closed form of the sequence.

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Activity 8.1.5.

Consider the sequence \(\displaystyle 1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots\text{.}\) Which of the following choices gives a closed formula for this sequence? Select all that apply.

\(\displaystyle \displaystyle \left\{\left(\frac{1}{3}\right)^{n-1}\right\}_{n=1}^\infty\)

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\(\displaystyle \displaystyle \left\{\left(\frac{1}{3}\right)^{n}\right\}_{n=1}^\infty\)

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\(\displaystyle \displaystyle \left\{\left(\frac{1}{3}\right)^{n-1}\right\}_{n=2}^\infty\)

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\(\displaystyle \displaystyle \left\{\left(\frac{1}{3}\right)^{n+1}\right\}_{n=0}^\infty\)

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\(\displaystyle \displaystyle \left\{\left(\frac{1}{3}\right)^{n}\right\}_{n=0}^\infty\)

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Activity 8.1.6.

Let \(a_n\) be the \(n\)th term in the sequence \(\displaystyle \left\{\frac{n+1}{n}\right\}_{n=1}^\infty\text{.}\) Which of the following terms corresponds to the \(27^{th}\) term of this sequence?

\(\displaystyle \frac{27}{26}\)

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\(\displaystyle \frac{26}{27}\)

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\(\displaystyle \frac{27}{28}\)

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\(\displaystyle \frac{28}{27}\)

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\(\displaystyle \frac{29}{28}\)

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Activity 8.1.7.

Let \(a_n\) be the \(n\)th term in the sequence \(\displaystyle \left\{\frac{n+1}{n}\right\}_{n=2}^\infty\text{.}\) Which of the following terms corresponds to the \(27^{th}\) term of this sequence?

\(\displaystyle \frac{27}{26}\)

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\(\displaystyle \frac{26}{27}\)

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\(\displaystyle \frac{27}{28}\)

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\(\displaystyle \frac{28}{27}\)

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\(\displaystyle \frac{29}{28}\)

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Activity 8.1.8.

Let \(a_n\) be the \(n\)th term in the sequence \(\displaystyle 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots\text{.}\) Identify the \(81\)st term of this sequence.

\(\displaystyle \frac{1}{79}\)

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\(\displaystyle \frac{1}{80}\)

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\(\displaystyle \frac{1}{81}\)

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\(\displaystyle \frac{1}{82}\)

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\(\displaystyle \frac{1}{83}\)

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Activity 8.1.9.

Find a closed form for the sequence \(0, 3, 8, 15, 24, \ldots\text{.}\)

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Activity 8.1.10.

Find a closed form for the sequence \(\displaystyle \frac{12}{1}, \frac{16}{2}, \frac{20}{3}, \frac{24}{4}, \frac{28}{5}, \ldots\text{.}\)

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Activity 8.1.11.

Let \(a_n\) be the \(n\)th term in the sequence \(1, 1, 2, 3, 5, 8, \ldots\text{.}\) Find a formula for \(a_n\text{.}\)

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Definition 8.1.12.

A sequence is recursive if the terms are defined as a function of previous terms (with the necessary initial terms provided).

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Activity 8.1.13.

Consider the sequence defined by \(a_1=6\) and \(a_{k+1}=4a_k-7\) for \(k\geq 1\text{.}\) What are the first four terms?

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Activity 8.1.14.

Consider the sequence \(2, 7, 22, 67, 202, \ldots\text{.}\) Which of the following offers the best recursive formula for this sequence?

\(\displaystyle a_{n+1} = 3a_n+1\)

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\(a_1=2, a_k=3a_{k-1}+1\) for \(k>1\)

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\(a_1=2, a_2=7, a_k=3a_{k-1}+1\) for \(k>2\)

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Activity 8.1.15.

Once more, consider the sequence \(1, 1, 2, 3, 5, 8, \ldots\) from Activity 8.1.11. Suppose \(a_1=1\) and \(a_2=1\text{.}\) Give a recursive formula for \(a_n\) for all \(n\geq 3\text{.}\)

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