Stan thinks there are more than \(200,000\) pennies in this pyramid. Do you agree with him? Justify your reasoning with mathematics.
Image Source: Dan Meyer 3 Acts |
A sequence is an ordered list of numbers. It is a function where the domain is a sequence of consecutive positive integers and the range is the terms of the sequence. The fact that the domain is a sequence of consecutive positive integers just means that you have a first term, second term, third term, etc.
A sequence can be finite or infinite. An example of a finite sequence could be \(2,\ 4,\ 6,\ 8,\ ...,\ 96,\ 98,\ 100\), and this is finite because it has a limited number of terms. An infinite sequence could be \(2,\ 4,\ 6,\ 8,\ 10,\ 12,\ ...\) which would continue without stopping.
We use subscripts to notate the terms: \(a_1\) represents the first term, \(a_2\) represents the second term, and \(a_n\) would represent the \(nth\) term in a sequence. This is similar to function notation, \(f(x)\), where \(x\) is the input and \(f(x)\) is the output. With sequences, \(n\) is the input and \(a_n\) is the output. For example, finding \(a_5\) and finding \(f(5)\) both mean that we are letting \(5\) be the input.
This first target will explore some of the general ideas of sequences and series with a focus on correct notation, and then we will explore specific types of sequences and series in future targets.
An explicit formula allows us to find any term in a sequence directly. For example, we can find the \(50th\) term by letting \(n=50\).
Example 1: Given the following formula: \(\large{a_n=\frac{n^3+2}{4n}}\), find the first five terms of the sequence.
To find the first term, we would let \(n=1\) and continue this for each subsequent term.
\(\large{a_1=\frac{1^3+2}{4\left(1\right)}=\frac{3}{4}}\)
\(\large{a_2=\frac{2^3+2}{4\left(2\right)}=\frac{5}{4}}\)
\(\large{a_3=\frac{3^3+2}{4\left(3\right)}=\frac{29}{12}}\)
\(\large{a_4=\frac{4^3+2}{4\left(4\right)}=\frac{33}{8}}\)
\(\large{a_5=\frac{5^3+2}{4\left(5\right)}=\frac{127}{20}}\)
So the first five terms of this sequence would be \(\large{\frac{3}{4},\ \frac{5}{4},\ \frac{29}{12},\ \frac{33}{8},\ \frac{127}{20}}\).
Example 2: Write the first \(5\) terms of the sequence: \(a(n) = \begin{cases} n^3\ \text{if}\ n\ \text{is even}\\ \frac{2}{n+5}\ \text{if}\ n\ \text{is odd} \\ \end{cases}\)
A sequence can be finite or infinite. An example of a finite sequence could be \(2,\ 4,\ 6,\ 8,\ ...,\ 96,\ 98,\ 100\), and this is finite because it has a limited number of terms. An infinite sequence could be \(2,\ 4,\ 6,\ 8,\ 10,\ 12,\ ...\) which would continue without stopping.
We use subscripts to notate the terms: \(a_1\) represents the first term, \(a_2\) represents the second term, and \(a_n\) would represent the \(nth\) term in a sequence. This is similar to function notation, \(f(x)\), where \(x\) is the input and \(f(x)\) is the output. With sequences, \(n\) is the input and \(a_n\) is the output. For example, finding \(a_5\) and finding \(f(5)\) both mean that we are letting \(5\) be the input.
This first target will explore some of the general ideas of sequences and series with a focus on correct notation, and then we will explore specific types of sequences and series in future targets.
An explicit formula allows us to find any term in a sequence directly. For example, we can find the \(50th\) term by letting \(n=50\).
Example 1: Given the following formula: \(\large{a_n=\frac{n^3+2}{4n}}\), find the first five terms of the sequence.
To find the first term, we would let \(n=1\) and continue this for each subsequent term.
\(\large{a_1=\frac{1^3+2}{4\left(1\right)}=\frac{3}{4}}\)
\(\large{a_2=\frac{2^3+2}{4\left(2\right)}=\frac{5}{4}}\)
\(\large{a_3=\frac{3^3+2}{4\left(3\right)}=\frac{29}{12}}\)
\(\large{a_4=\frac{4^3+2}{4\left(4\right)}=\frac{33}{8}}\)
\(\large{a_5=\frac{5^3+2}{4\left(5\right)}=\frac{127}{20}}\)
So the first five terms of this sequence would be \(\large{\frac{3}{4},\ \frac{5}{4},\ \frac{29}{12},\ \frac{33}{8},\ \frac{127}{20}}\).
Example 2: Write the first \(5\) terms of the sequence: \(a(n) = \begin{cases} n^3\ \text{if}\ n\ \text{is even}\\ \frac{2}{n+5}\ \text{if}\ n\ \text{is odd} \\ \end{cases}\)
The sequence we worked with above, \(a_n=\frac{n^3+2}{4n}\), follows a specific pattern, but it would be difficult to come up with that formula based on the terms alone. On the other hand, some sequences follow an easily recognizable pattern.
Example 3: Determine the sequence from the pattern.
a) \(5,\ \large{\frac{5}{6},\ \frac{5}{36},\ \frac{5}{216},...}\) b) \(1,\ \sqrt{2},\ \sqrt{3},\ 2,\ ...\) c) \(5,\ -3,\ -11,\ -19,\ ...\)
A series is the expression formed when we take the terms in a sequence and add them up. We can have a finite series (for example, adding up the first ten terms in a sequence) or we can have an infinite series where we add up an infinite number of terms in a sequence (we will explore this more in a later target).
We will use the Greek letter Sigma (\(\Sigma\)) for the notation, and it is often referred to as summation notation or Sigma notation. The word “sigma” means to add up or find a mathematical sum.
Summation Notation: \(\large\sum\limits_{k=1}^{n}{a_k}\)
\(k\) is the index and the lower number (in this case, \(1\)), tells us that we are starting our sum with the first term in the sequence, and the upper number (in this case, \(n\)) tells us that we will add up all of the terms and stop at the \(nth\) term.
\(k\) is the index and the lower number (in this case, \(1\)), tells us that we are starting our sum with the first term in the sequence, and the upper number (in this case, \(n\)) tells us that we will add up all of the terms and stop at the \(nth\) term.
Here is an example: \(\large\sum\limits_{k=1}^{5}{k^2}\)
This notation tells us that we are adding the first through fifth terms of the sequence \(a_k=k^2\) so, \(\large\sum\limits_{k=1}^{5}{k^2}=1^2+2^2+3^2+4^2+5^2\).
Example 4: Express the sum \(\frac{x}{3}+\frac{x^2}{6}+\frac{x^3}{9}+...+\frac{x^{10}}{30}\) in summation notation.