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Section 1.4 Linear and Exponential Growth

Let \(b,m\) be real constants, and consider the linear function \(L(t)=b+mt\text{.}\) The sequence of values \(L(0), L(1), L(2),\ldots\) given by
\begin{equation*} b, b+m, b+2m,\ldots ,b+nm,\ldots \end{equation*}
is called an arithmetic sequence
The emphasis is on the third syllable "met" when the word "arithmetic" is used as an adjective rather than a noun. For example: "Addition is an operation of \(\mbox{a}\cdot \mbox{rith}'\cdot \mbox{metic}\text{.}\) Repeated addition creates an \(\mbox{arith}\cdot \mbox{met}'\cdot \mbox{ic}\) sequence."
with initial term \(b\) and common difference \(m\text{.}\) An arithmetic sequence is said to exhibit linear growth or decay, according to whether \(m \gt 0\) or \(m\lt 0\text{,}\) respectively.
Let \(a,r\) be real constants with \(a\neq 0, r\gt 0, r\neq 1\text{,}\) and consider the exponential function \(E(t) = ar^t\text{.}\) The sequence of values \(E(0),E(1),E(2),\ldots\) given by
\begin{equation*} a,ar,ar^2,\ldots,ar^n,\ldots \end{equation*}
is called a geometric sequence with initial term \(a\) and common ratio \(r\text{.}\) A geometric sequence with \(a\gt 0\) is said to exhibit exponential growth or decay, according to whether \(r\gt 1\) or \(r\lt 1\text{,}\) respectively.

Checkpoint 1.4.1.

Fill in the missing terms of the following arithmetic and geometric sequences. Identify the initial term and the common difference or common ratio for each.
  1. \(\displaystyle 5,2,-1, \underline{\rule{2ex}{0ex}}, \underline{\rule{2ex}{0ex}},\underline{\rule{2ex}{0ex}},\ldots\)
  2. \(\displaystyle 5,2,0.8,\underline{\rule{2ex}{0ex}}, \underline{\rule{2ex}{0ex}},\underline{\rule{2ex}{0ex}},\ldots\)
  3. \(\displaystyle \underline{\rule{2ex}{0ex}}, 2, \underline{\rule{2ex}{0ex}},5,\underline{\rule{2ex}{0ex}},8,\ldots\)
  4. \(\displaystyle \underline{\rule{2ex}{0ex}}, 2, \underline{\rule{2ex}{0ex}},4,\underline{\rule{2ex}{0ex}},8,\ldots\)
Finite arithmetic and geometric sums. Exercises at the end of this subsection outline the derivations of the following formulas.
\begin{align} b + (b+m) + (b+2m) + \cdots + (b+nm) \amp = \frac{(n+1)(2b+nm)}{2} \tag{1.4.1}\\ a + ar + ar^2 + \cdots + ar^n \amp = a\left(\frac{1-r^{n+1}}{1-r}\right)\tag{1.4.2} \end{align}

Checkpoint 1.4.2.

Find the given sums of terms of arithmetic and geometric sequences.
  1. Find the sum of the first 100 positive integers.
  2. \(\displaystyle 2 + 5 + 8 + 11 + \cdots + 302\)
  3. \(\displaystyle 2 + 5 + 8 + 11 + \cdots + 1571\)
  4. \(\displaystyle 2 + 6 + 18 + 54 + \cdots + 2(3^{100})\)
  5. \(\displaystyle 2 + 6 + 18 + 54 + \cdots + 9565938\)
Infinite geometric sums. An infinite sum of the form
\begin{equation*} a + ar + ar^2 + ar^3 + \cdots \end{equation*}
is called an infinite geometric series, and is defined to mean \(\displaystyle \lim_{n\to\infty} s_n\) (if the limit exists), where \(s_1,s_2,s_3,\ldots\) is sequence of finite sums
\begin{align*} s_0 \amp = a\\ s_1 \amp = a + ar\\ s_2 \amp = a + ar + ar^2\\ \amp \vdots \\ s_n \amp = a+ ar+ar^2 + \cdots + ar^n\\ \amp \vdots \end{align*}
If \(|r| \lt 1\text{,}\) then \(|r|^n \to 0\) as \(n\to \infty\text{.}\) Using properties of limits from calculus, we have
\begin{equation*} a\left(\frac{1-r^{n+1}}{1-r} \right) \to a\left(\frac{1}{1-r}\right) \end{equation*}
as \(n\to \infty\text{.}\) Putting this together with (1.4.2) above is the justification for the following formula.
\begin{align} a + ar+ ar^2 + ar^3 + \cdots = a\left(\frac{1}{1-r}\right) \amp \amp \mbox{ for } |r|\lt 1\tag{1.4.3} \end{align}

Exercises Exercises


Derive (1.4.1).
Write the sum in reverse order \(L(n)+L(n-1) + \cdots + L(1) + L(0)\) directly beneath \(L(0)+ L(1) + \cdots + L(n)\text{,}\) in such a way that the terms are aligned vertically. Notice that each vertically aligned pair has the form \(L(k)\) and \(L(n-k)\text{,}\) and that \(L(k)+L(n-k) = 2b+nm\) (the \(k\)’s cancel!). Now go from there.


Derive (1.4.2).
Let \(s\) be the desired sum \(a+ar+ar^2 + \cdots + ar^n\text{.}\) Examine the expansion of \(s-rs\) (many terms cancel!). Simplify and solve for \(s\text{.}\)