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Sequences

Write the first few terms of a sequence

Write the First Few Terms of a Sequence

Let’s look at the function \(f(x)=2x\) and evaluate it for just the counting numbers.

\(f(x)=2x\)
\(x\)	\(2x\)
1	2
2	4
3	6
4	8
5	10
…	…

If we list the function values in order as 2, 4, 6, 8, and 10, … we have a sequence. A sequence is a function whose domain is the counting numbers.

A sequence can also be seen as an ordered list of numbers and each number in the list is a term. A sequence may have an infinite number of terms or a finite number of terms. Our sequence has three dots (ellipsis) at the end which indicates the list never ends. If the domain is the set of all counting numbers, then the sequence is an infinite sequence. Its domain is all counting numbers and there is an infinite number of counting numbers.

\[2,4,6,8,10,\text{\ldots },\]

If we limit the domain to a finite number of counting numbers, then the sequence is a finite sequence. If we use only the first four counting numbers, 1, 2, 3, 4 our sequence would be the finite sequence,

\[2,4,6,8\]

Often when working with sequences we do not want to write out all the terms. We want more compact way to show how each term is defined. When we worked with functions, we wrote \(f(x)=2x\) and we said the expression \(2x\) was the rule that defined values in the range. While a sequence is a function, we do not use the usual function notation. Instead of writing the function as \(f(x)=2x,\) we would write it as \({a}_{n}=2n.\) The \({a}_{n}\) is the nth term of the sequence, the term in the nth position where n is a value in the domain. The formula for writing the nth term of the sequence is called the general term or formula of the sequence.

When we are given the general term of the sequence, we can find the terms by replacing n with the counting numbers in order. For \({a}_{n}=2n,\)

\(n\)	1	2	3	4	5	\({a}_{n}\)
\({a}_{n}\)	\(\begin{array}{l}2\cdot 1 \\ 2\end{array}\)	\(\begin{array}{l}2\cdot 2 \\ 4\end{array}\)	\(\begin{array}{l}2\cdot 3 \\ 6\end{array}\)	\(\begin{array}{l}2\cdot 4 \\ 8\end{array}\)	\(\begin{array}{l}2\cdot 5 \\ 10\end{array}\)	2n

\[\begin{array}{llllll}{a}_{1}, & {a}_{2}, & {a}_{3}, & {a}_{4}, & {a}_{5}\text{,}\ \text{\ldots }\text{,} & {a}_{n}\text{,}\ \text{\ldots } \\ 2, & 4, & 6, & 8, & 10\text{,}\ \text{\ldots } & \end{array}\]

To find the values of a sequence, we substitute in the counting numbers in order into the general term of the sequence.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Find a Formula for the General Term (

Sometimes we have a few terms of a sequence and it would be helpful to know the general term or nth term. To find the general term, we look for patterns in the terms. Often the patterns involve multiples or powers. We also look for a pattern in the signs of the terms.

Example

Try it.

Find a general term for the sequence whose first five terms are shown.

\[4,8,12,16,20\text{,}\ \ldots\]

Solution



We look for a pattern in the terms.
The numbers are all multiples of 4.
	The general term of the sequence is \({a}_{n}=4n.\)

Example

Try it.

Find a general term for the sequence whose first five terms are shown.

\[2,-4,8,-16,32,\text{\ldots }\]

Solution



We look for a pattern in the terms.
The numbers are powers of 2. The signs are alternating, with even \(n\) negative.
	The general term of the sequence is \({a}_{n}={(-1)}^{n+1}{2}^{n}.\)

Example

Try it.

Find a general term for the sequence whose first five terms are shown.

\[\frac{1}{3},\frac{1}{9},\frac{1}{27},\frac{1}{81},\frac{1}{243},\text{\ldots }\]

Solution



We look for a pattern in the terms.
The numerators are all 1.
The denominators are powers of 3.	The general term of the sequence is \({a}_{n}=\frac{1}{{3}^{n}}.\)

Use Factorial Notation

Sequences often have terms that are products of consecutive integers. We indicate these products with a special notation called factorial notation. For example,\(5!\), read 5 factorial, means \(5\cdot 4\cdot 3\cdot 2\cdot 1.\) The exclamation point is not punctuation here; it indicates the factorial notation.

The values of \(n!\) for the first 5 positive integers are shown.

\[\begin{array}{lllllllllllll}1! & & & \ 2! & & & \ 3! & & & \ 4! & & & \ 5! \\ 1 & & & \ 2⋅1 & & & \ 3⋅2⋅1 & & & \ 4⋅3⋅2⋅1 & & & \ 5⋅4⋅3⋅2⋅1 \\ 1 & & & \ 2 & & & \ 6 & & & \ 24 & & & \ 120\end{array}\]

Example

Try it.

Write the first five terms of the sequence whose general term is \({a}_{n}=\frac{1}{n!}\).

Solution

We substitute the values 1, 2, 3, 4, 5 into the formula, \({a}_{n}=\frac{1}{n!},\) in order.

The first five terms of the sequence are \(1,\frac{1}{2},\frac{1}{6},\frac{1}{24},\frac{1}{120}.\)

When there is a fraction with factorials in the numerator and denominator, we line up the factors vertically to make our calculations easier.

Example

Try it.

Write the first five terms of the sequence whose general term is \({a}_{n}=\frac{(n+1)!}{(n-1)!}.\)

Solution

We substitute the values 1, 2, 3, 4, 5 into the formula, \({a}_{n}=\frac{(n+1)!}{(n-1)!},\) in order.

The first five terms of the sequence are 2, 6, 12, 20, and 30.

Find the Partial Sum

Sometimes in applications, rather than just list the terms, it is important for us to add the terms of a sequence. Rather than just connect the terms with plus signs, we can use summation notation.

For example, \({a}_{1}+{a}_{2}+{a}_{3}+{a}_{4}+{a}_{5}\) can be written as \(\sum _{i=1}^{5}{a}_{i}.\) We read this as “the sum of a sub i from i equals one to five.” The symbol \(\sum\) means to add and the i is the index of summation. The 1 tells us where to start (initial value) and the 5 tells us where to end (terminal value).

When we add a finite number of terms, we call the sum a partial sum.

Example

Try it.

Expand the partial sum and find its value: \(\sum _{i=1}^{5}2i.\)

Solution

	\(\sum _{i=1}^{5}2i\)
We substitute the values 1, 2, 3, 4, 5 in order.	\(2\cdot 1+2\cdot 2+2\cdot 3+2\cdot 4+2\cdot 5\)
Simplify.	\(2+4+6+8+10\)
Add.	\(30\)
	\(\sum _{i=1}^{5}2i=30\)

The index does not always have to be i we can use any letter, but i and k are commonly used. The index does not have to start with 1 either—it can start and end with any positive integer.

Example

Try it.

Expand the partial sum and find its value: \(\sum _{k=0}^{3}\frac{1}{k!}.\)

Solution

	\(\sum _{k=0}^{3}\frac{1}{k!}\)
We substitute the values 0, 1, 2, 3, in order.	\(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}\)
Evaluate the factorials.	\(\frac{1}{1}+\frac{1}{1}+\frac{1}{2}+\frac{1}{6}\)
Simplify.	\(1+1+\frac{3}{6}+\frac{1}{6}\)
Simplify.	\(\frac{16}{6}\)
Simplify.	\(\frac{8}{3}\)
	\(\sum _{k=0}^{3}\frac{1}{k!}=\frac{8}{3}\)

Use Summation Notation to Write a Sum

In the last two examples, we went from summation notation to writing out the sum. Now we will start with a sum and change it to summation notation. This is very similar to finding the general term of a sequence. We will need to look at the terms and find a pattern. Often the patterns involve multiples or powers.

Example

Try it.

Write the sum using summation notation: \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}.\)

Solution

	\(\ 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\)
	\(\ n\text{:}\ 1,2,3,4,5\)
We look for a pattern in the terms.	Terms: \(1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5}\)
The numerators are all one.	Pattern: \(\frac{1}{1},\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\ldots \frac{1}{n}\)
The denominators are the counting numbers from one to five.	The sum written in summation notation is
	\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\sum _{n=1}^{5}\frac{1}{n}.\)

When the terms of a sum have negative coefficients, we must carefully analyze the pattern of the signs.

Example

Try it.

Write the sum using summation notation: \(-1+8-27+64-125.\)

Solution



We look for a pattern in the terms.
The signs of the terms alternate, and the odd terms are negative.
The numbers are the cubes of the counting numbers from one to five.

	The sum written in summation notation is
	\(-1+8-27+64-125=\sum _{n=1}^{5}{(-1)}^{n}⋅{n}^{3}\)

Key Concepts

Factorial Notation
If n is a positive integer, then \(n!\) is
\[n!=n(n-1)(n-2)\ldots (3)(2)(1).\]
We define \(0!\) as 1, so \(0!=1\)
Summation Notation
The sum of the first n terms of a sequence whose nth term \({a}_{n}\) is written in summation notation as:
\[\sum _{i=1}^{n}{a}_{i}={a}_{1}+{a}_{2}+{a}_{3}+{a}_{4}+{a}_{5}+\ldots +{a}_{n}\]
The i is the index of summation and the 1 tells us where to start and the n tells us where to end.

Sequences

Write the First Few Terms of a Sequence

In the following exercises, write the first five terms of the sequence whose general term is given.

Try it.

\({a}_{n}=2n-7\)

Solution

\(-5,-3,-1,1,3\)

Try it.

\({a}_{n}=5n-1\)

Try it.

\({a}_{n}=3n+1\)

Solution

\(4,7,10,13,16\)

Try it.

\({a}_{n}=4n+2\)

Try it.

\({a}_{n}={2}^{n}+3\)

Solution

\(5,7,11,19,35\)

Try it.

\({a}_{n}={3}^{n}-1\)

Try it.

\({a}_{n}={3}^{n}-2n\)

Solution

\(1,5,21,73,233\)

Try it.

\({a}_{n}={2}^{n}-3n\)

Try it.

\({a}_{n}=\frac{{2}^{n}}{{n}^{2}}\)

Solution

\(2,1,\frac{8}{9},1,\frac{32}{25}\)

Try it.

\({a}_{n}=\frac{{3}^{n}}{{n}^{3}}\)

Try it.

\({a}_{n}=\frac{4n-2}{{2}^{n}}\)

Solution

\(1,\frac{3}{2},\frac{5}{4},\frac{7}{8},\frac{9}{16}\)

Try it.

\({a}_{n}=\frac{3n+3}{{3}^{n}}\)

Try it.

\({a}_{n}={(-1)}^{n}\cdot 2n\)

Solution

\(-2,4,-6,8,-10\)

Try it.

\({a}_{n}={(-1)}^{n}\cdot 3n\)

Try it.

\({a}_{n}={(-1)}^{n+1}{n}^{2}\)

Solution

\(1,-4,9,-16,25\)

Try it.

\({a}_{n}={(-1)}^{n+1}{n}^{4}\)

Try it.

\({a}_{n}=\frac{{(-1)}^{n+1}}{{n}^{2}}\)

Solution

\(1,-\frac{1}{4},\frac{1}{9},-\frac{1}{16},\frac{1}{25}\)

Try it.

\({a}_{n}=\frac{{(-1)}^{n+1}}{2n}\)

Find a Formula for the General Term (nth Term) of a Sequence

In the following exercises, find a general term for the sequence whose first five terms are shown.

Try it.

\(8,16,24,32,40,\text{\ldots }\)

Solution

\({a}_{n}=8n\)

Try it.

\(7,14,21,28,35,\text{\ldots }\)

Try it.

\(6,7,8,9,10,\text{\ldots }\)

Solution

\({a}_{n}=n+5\)

Try it.

\(-3,-2,-1,0,1,\text{\ldots }\)

Try it.

\({e}^{3},{e}^{4},{e}^{5},{e}^{6},{e}^{7},\text{\ldots }\)

Solution

\({a}_{n}={e}^{n+2}\)

Try it.

\(\frac{1}{{e}^{2}},\frac{1}{e},1,e,{e}^{2},\text{\ldots }\)

Try it.

\(-5,10,-15,20,-25,\text{\ldots }\)

Solution

\({a}_{n}={(-1)}^{n}5n\)

Try it.

\(-6,11,-16,21,-26,\text{\ldots }\)

Try it.

\(-1,8,-27,64,-125,\text{\ldots }\)

Solution

\({a}_{n}={(-1)}^{n}{n}^{3}\)

Try it.

\(2,-5,10,-17,26,\text{\ldots }\)

Try it.

\(-2,4,-6,8,-10,\text{\ldots }\)

Solution

\({a}_{n}={(-1)}^{n}2n\)

Try it.

\(1,-3,5,-7,9,\text{\ldots }\)

Try it.

\(\frac{1}{4},\frac{1}{16},\frac{1}{64},\frac{1}{256},\frac{1}{1,024},\text{\ldots }\)

Solution

\({a}_{n}=\frac{1}{{4}^{n}}\)

Try it.

\(\frac{1}{1},\frac{1}{8},\frac{1}{27},\frac{1}{64},\frac{1}{125},\text{\ldots }\)

Try it.

\(-\frac{1}{2},-\frac{2}{3},-\frac{3}{4},-\frac{4}{5},-\frac{5}{6},\text{\ldots }\)

Solution

\({a}_{n}=-\frac{n}{n+1}\)

Try it.

\(-2,-\frac{3}{2},-\frac{4}{3},-\frac{5}{4},-\frac{6}{5},\text{\ldots }\)

Try it.

\(-\frac{5}{2},-\frac{5}{4},-\frac{5}{8},-\frac{5}{16},-\frac{5}{32},\text{\ldots }\)

Solution

\({a}_{n}=-\frac{5}{{2}^{n}}\)

Try it.

\(4,\frac{1}{2},\frac{4}{27},\frac{4}{64},\frac{4}{125},\text{\ldots }\)

Use Factorial Notation

In the following exercises, using factorial notation, write the first five terms of the sequence whose general term is given.

Try it.

\({a}_{n}=\frac{4}{n!}\)

Solution

\(4,2,\frac{2}{3},\frac{1}{6},\frac{1}{30}\)

Try it.

\({a}_{n}=\frac{5}{n!}\)

Try it.

\({a}_{n}=3n!\)

Solution

\(3,6,18,72,360\)

Try it.

\({a}_{n}=2n!\)

Try it.

\({a}_{n}=(2n)!\)

Solution

\(2,24,720,40320,3628800\)

Try it.

\({a}_{n}=(3n)!\)

Try it.

\({a}_{n}=\frac{(n-1)!}{(n)!}\)

Solution

\(1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5}\)

Try it.

\({a}_{n}=\frac{n!}{(n+1)!}\)

Try it.

\({a}_{n}=\frac{n!}{{n}^{\text{2}}}\)

Solution

\(1,\frac{1}{2},\frac{2}{3},\frac{3}{2},\frac{24}{5}\)

Try it.

\({a}_{n}=\frac{{n}^{\text{2}}}{n!}\)

Try it.

\({a}_{n}=\frac{(n+1)!}{{n}^{2}}\)

Solution

\(2,\frac{3}{2},\frac{8}{3},\frac{15}{2},\frac{144}{5}\)

Try it.

\({a}_{n}=\frac{(n+1)!}{2n}\)

Find the Partial Sum

In the following exercises, expand the partial sum and find its value.

Try it.

\(\sum _{i=1}^{5}{i}^{2}\)

Solution

\(1+4+9+16+25=55\)

Try it.

\(\sum _{i=1}^{5}{i}^{3}\)

Try it.

\(\sum _{i=1}^{6}(2i+3)\)

Solution

\(5+7+9+11+13+15=60\)

Try it.

\(\sum _{i=1}^{6}(3i-2)\)

Try it.

\(\sum _{i=1}^{4}{2}^{i}\)

Solution

\(2+4+8+16=30\)

Try it.

\(\sum _{i=1}^{4}{3}^{i}\)

Try it.

\(\sum _{k=0}^{3}\frac{4}{k!}\)

Solution

\(\frac{4}{1}+\frac{4}{1}+\frac{4}{2}+\frac{4}{6}=\frac{32}{3}=10\frac{2}{3}\)

Try it.

\(\sum _{k=0}^{4}-\frac{1}{k!}\)

Try it.

\(\sum _{k=1}^{5}k(k+1)\)

Solution

\(2+6+12+20+30=70\)

Try it.

\(\sum _{k=1}^{5}k(2k-3)\)

Try it.

\(\sum _{n=1}^{5}\frac{n}{n+1}\)

Solution

\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+\frac{5}{6}=\frac{71}{20}\)

Try it.

\(\sum _{n=1}^{4}\frac{n}{n+2}\)

Use Summation Notation to write a Sum

In the following exercises, write each sum using summation notation.

Try it.

\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

Solution

\(\sum _{n=1}^{5}\frac{1}{{3}^{n}}\)

Try it.

\(\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\frac{1}{256}\)

Try it.

\(1+\frac{1}{8}+\frac{1}{27}+\frac{1}{64}+\frac{1}{125}\)

Solution

\(\sum _{n=1}^{5}\frac{1}{{n}^{3}}\)

Try it.

\(\frac{1}{5}+\frac{1}{25}+\frac{1}{125}+\frac{1}{625}\)

Try it.

\(2+1+\frac{2}{3}+\frac{1}{2}+\frac{2}{5}\)

Solution

\(\sum _{n=1}^{5}\frac{2}{n}\)

Try it.

\(3+\frac{3}{2}+1+\frac{3}{4}+\frac{3}{5}+\frac{1}{2}\)

Try it.

\(3-6+9-12+15\)

Solution

\(\sum _{n=1}^{5}{(-1)}^{n+1}3n\)

Try it.

\(-5+10-15+20-25\)

Try it.

\(-2+4-6+8-10+\ldots +20\)

Solution

\(\sum _{n=1}^{10}{(-1)}^{n}2n\)

Try it.

\(1-3+5-7+9+\ldots +21\)

Try it.

\(14+16+18+20+22+24+26\)

Solution

\(\sum _{n=1}^{7}(2n+12)\)

Try it.

\(9+11+13+15+17+19+21\)

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Evaluate \(2n+3\) for the integers 1, 2, 3, and 4.
If you missed this problem, review .

जवाफ प्रकट गर्नुहोस्

5, 7, 9, 11
Evaluate \({(-1)}^{n}\) for the integers 1, 2, 3, and 4.
If you missed this problem, review .

जवाफ प्रकट गर्नुहोस्

\(-1,1,-1,1\)
If \(f(n)={n}^{2}+2,\) find \(f(1)+f(2)+f(3).\)
If you missed this problem, review .

जवाफ प्रकट गर्नुहोस्

20
Write the first five terms of the sequence whose general term is \({a}_{n}=4n-3.\)

जवाफ प्रकट गर्नुहोस्

We substitute the values 1, 2, 3, 4, and 5 into the formula, \({a}_{n}=4n-3,\) in order.
The first five terms of the sequence are 1, 5, 9, 13, and 17.
Write the first five terms of the sequence whose general term is \({a}_{n}=3n-4.\)

जवाफ प्रकट गर्नुहोस्

\(-1,2,5,8,11\)
Write the first five terms of the sequence whose general term is \({a}_{n}=2n-5.\)

जवाफ प्रकट गर्नुहोस्

\(-3,-1,1,3,5\)
Write the first five terms of the sequence whose general term is \({a}_{n}={2}^{n}+1.\)

जवाफ प्रकट गर्नुहोस्

We substitute the values 1, 2, 3, 4, and 5 into the formula, \({a}_{n}={2}^{n}+1,\) in order.
The first five terms of the sequence are 3, 5, 9, 17, and 33.
Write the first five terms of the sequence whose general term is \({a}_{n}={3}^{n}+4.\)

जवाफ प्रकट गर्नुहोस्

\(7,13,31,85,247\)
Write the first five terms of the sequence whose general term is \({a}_{n}={2}^{n}-5.\)

जवाफ प्रकट गर्नुहोस्

\(-3,-1,3,11,27\)
Write the first five terms of the sequence whose general term is \({a}_{n}={(-1)}^{n}{n}^{3}.\)

जवाफ प्रकट गर्नुहोस्

We substitute the values 1, 2, 3, 4, and 5 into the formula, \({a}_{n}={(-1)}^{n}{n}^{3},\) in order.
The first five terms of the sequence are \(-1,8,-27,64,\) and \(-125.\)
Write the first five terms of the sequence whose general term is \({a}_{n}={(-1)}^{n}{n}^{2}.\)

जवाफ प्रकट गर्नुहोस्

\(-1,4,-9,16,-25\)
Write the first five terms of the sequence whose general term is \({a}_{n}={(-1)}^{n+1}{n}^{3}.\)

जवाफ प्रकट गर्नुहोस्

\(1,-8,27,-64,125\)
Find a general term for the sequence whose first five terms are shown.
\[4,8,12,16,20\text{,}\ \ldots\]

जवाफ प्रकट गर्नुहोस्

We look for a pattern in the terms.
The numbers are all multiples of 4.
The general term of the sequence is \({a}_{n}=4n.\)
Find a general term for the sequence whose first five terms are shown.
\(3,6,9,12,15,\text{\ldots }\)

जवाफ प्रकट गर्नुहोस्

\({a}_{n}=3n\)
Find a general term for the sequence whose first five terms are shown.
\(5,10,15,20,25,\text{\ldots }\)

जवाफ प्रकट गर्नुहोस्

\({a}_{n}=5n\)
Find a general term for the sequence whose first five terms are shown.
\[2,-4,8,-16,32,\text{\ldots }\]

जवाफ प्रकट गर्नुहोस्

We look for a pattern in the terms.
The numbers are powers of 2. The signs are
alternating, with even \(n\) negative.
The general term of the sequence is \({a}_{n}={(-1)}^{n+1}{2}^{n}.\)
Find a general term for the sequence whose first five terms are shown.
\(-3,9,-27,81,-243\text{,}\ \text{\ldots }\)

जवाफ प्रकट गर्नुहोस्

\({a}_{n}={(-1)}^{n}{3}^{n}\)
Find a general term for the sequence whose first five terms are shown
\(1,-4,9,-16,25\text{,}\ \text{\ldots }\)

जवाफ प्रकट गर्नुहोस्

\({a}_{n}={(-1)}^{n+1}{n}^{2}\)
Find a general term for the sequence whose first five terms are shown.
\[\frac{1}{3},\frac{1}{9},\frac{1}{27},\frac{1}{81},\frac{1}{243},\text{\ldots }\]

जवाफ प्रकट गर्नुहोस्

We look for a pattern in the terms.
The numerators are all 1.
The denominators are powers of 3. The general term of the sequence is \({a}_{n}=\frac{1}{{3}^{n}}.\)
Find a general term for the sequence whose first five terms are shown.
\(\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},\frac{1}{32},\text{\ldots }\)

जवाफ प्रकट गर्नुहोस्

\({a}_{n}=\frac{1}{{2}^{n}}\)
Find a general term for the sequence whose first five terms are shown.
\(\frac{1}{1},\frac{1}{4},\frac{1}{9},\frac{1}{16},\frac{1}{25},\text{\ldots }\)

जवाफ प्रकट गर्नुहोस्

\({a}_{n}=\frac{1}{{n}^{2}}\)
Write the first five terms of the sequence whose general term is \({a}_{n}=\frac{1}{n!}\).

जवाफ प्रकट गर्नुहोस्

We substitute the values 1, 2, 3, 4, 5 into the formula, \({a}_{n}=\frac{1}{n!},\) in order.
The first five terms of the sequence are \(1,\frac{1}{2},\frac{1}{6},\frac{1}{24},\frac{1}{120}.\)
Write the first five terms of the sequence whose general term is \({a}_{n}=\frac{2}{n!}.\)

जवाफ प्रकट गर्नुहोस्

\(2,1,\frac{1}{3},\frac{1}{12},\frac{1}{60}\)
Write the first five terms of the sequence whose general term is \({a}_{n}=\frac{3}{n!}.\)

जवाफ प्रकट गर्नुहोस्

\(3,\frac{3}{2},\frac{1}{2},\frac{1}{8},\frac{1}{40}\)
Write the first five terms of the sequence whose general term is \({a}_{n}=\frac{(n+1)!}{(n-1)!}.\)

जवाफ प्रकट गर्नुहोस्

We substitute the values 1, 2, 3, 4, 5 into the formula, \({a}_{n}=\frac{(n+1)!}{(n-1)!},\) in order.
The first five terms of the sequence are 2, 6, 12, 20, and 30.
Write the first five terms of the sequence whose general term is \({a}_{n}=\frac{(n-1)!}{(n+1)!}.\)

जवाफ प्रकट गर्नुहोस्

\(\frac{1}{2},\frac{1}{6},\frac{1}{12},\frac{1}{20},\frac{1}{30}\)
Write the first five terms of the sequence whose general term is \({a}_{n}=\frac{n!}{(n+1)!}.\)

जवाफ प्रकट गर्नुहोस्

\(\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6}\)
Expand the partial sum and find its value: \(\sum _{i=1}^{5}2i.\)

जवाफ प्रकट गर्नुहोस्

\(\sum _{i=1}^{5}2i\)
We substitute the values 1, 2, 3, 4, 5 in order. \(2\cdot 1+2\cdot 2+2\cdot 3+2\cdot 4+2\cdot 5\)
Simplify. \(2+4+6+8+10\)
Add. \(30\)
\(\sum _{i=1}^{5}2i=30\)
Expand the partial sum and find its value: \(\sum _{i=1}^{5}3i.\)

जवाफ प्रकट गर्नुहोस्

45
Expand the partial sum and find its value: \(\sum _{i=1}^{5}4i.\)

जवाफ प्रकट गर्नुहोस्

60

Expand the partial sum and find its value: \(\sum _{k=0}^{3}\frac{1}{k!}.\)

जवाफ प्रकट गर्नुहोस्

	\(\sum _{k=0}^{3}\frac{1}{k!}\)
We substitute the values 0, 1, 2, 3, in order.	\(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}\)
Evaluate the factorials.	\(\frac{1}{1}+\frac{1}{1}+\frac{1}{2}+\frac{1}{6}\)
Simplify.	\(1+1+\frac{3}{6}+\frac{1}{6}\)
Simplify.	\(\frac{16}{6}\)
Simplify.	\(\frac{8}{3}\)
	\(\sum _{k=0}^{3}\frac{1}{k!}=\frac{8}{3}\)

Expand the partial sum and find its value: \(\sum _{k=0}^{3}\frac{2}{k!}.\)

जवाफ प्रकट गर्नुहोस्

\(\frac{16}{3}\)
Expand the partial sum and find its value: \(\sum _{k=0}^{3}\frac{3}{k!}.\)

जवाफ प्रकट गर्नुहोस्

8

Write the sum using summation notation: \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}.\)

जवाफ प्रकट गर्नुहोस्

	\(\ 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\)
	\(\ n\text{:}\ 1,2,3,4,5\)
We look for a pattern in the terms.	Terms: \(1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5}\)
The numerators are all one.	Pattern: \(\frac{1}{1},\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\ldots \frac{1}{n}\)
The denominators are the counting numbers from one to five.	The sum written in summation notation is
	\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\sum _{n=1}^{5}\frac{1}{n}.\)

Write the sum using summation notation: \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}.\)

जवाफ प्रकट गर्नुहोस्

\(\sum _{n=1}^{5}\frac{1}{{2}^{n}}\)
Write the sum using summation notation: \(1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}.\)

जवाफ प्रकट गर्नुहोस्

\(\sum _{n=1}^{5}\frac{1}{{n}^{2}}\)

Write the sum using summation notation: \(-1+8-27+64-125.\)

जवाफ प्रकट गर्नुहोस्



We look for a pattern in the terms.
The signs of the terms alternate, and the odd terms are negative.
The numbers are the cubes of the counting numbers from one to five.

	The sum written in summation notation is
	\(-1+8-27+64-125=\sum _{n=1}^{5}{(-1)}^{n}⋅{n}^{3}\)

Write each sum using summation notation: \(1-4+9-16+25.\)

जवाफ प्रकट गर्नुहोस्

\(\sum _{n=1}^{5}{(-1)}^{n+1}{n}^{2}\)
Write each sum using summation notation: \(-2+4-6+8-10.\)

जवाफ प्रकट गर्नुहोस्

\(\sum _{n=1}^{5}{(-1)}^{n}2n\)
\({a}_{n}=2n-7\)

जवाफ प्रकट गर्नुहोस्

\(-5,-3,-1,1,3\)

Symbols used here

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

factorial
n × (n−1) × … × 1; the number of orderings of n things. 0! = 1.

imaginary unit
i² = −1.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Sequences

Write the first few terms of a sequence
Find a formula for the general term (nth term) of a sequence
Use factorial notation
Find the partial sum
Use summation notation to write a sum

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

तपाईँको आफ्नै प्रयास गर्नुहोस्

Parts of this page are adapted from OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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