Chapter 13
Algebra and Trigonometry · 296 exercises
Problem 1
Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 2+4+6+\cdots+2 n=n(n+1) $$
5 step solution
Problem 1
For the function \(f(x)=\frac{x-1}{x},\) find \(f(2)\) and \(f(3)\)
5 step solution
Problem 2
Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 1+5+9+\cdots+(4 n-3)=n(2 n-1) $$
4 step solution
Problem 2
How much do you need to invest now at \(5 \%\) per annum compounded monthly so that in 1 year you will have $$\$ 10,000 ?
7 step solution
Problem 2
True or False A function is a relation between two sets \(D\) and \(R\) so that each element \(x\) in the first set \(D\) is related to exactly one element \(y\) in the second set \(R\)
5 step solution
Problem 3
Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 3+4+5+\cdots+(n+2)=\frac{1}{2} n(n+5) $$
9 step solution
Problem 4
Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 3+5+7+\cdots+(2 n+1)=n(n+2) $$
6 step solution
Problem 4
If \(|r|<1,\) the sum of the geometric series \(\sum_{k=1}^{\infty} a r^{k-1}\) is _____ .
4 step solution
Problem 4
True or False The sum \(S_{n}\) of the first \(n\) terms of an arithmetic sequence \(\left\\{a_{n}\right\\}\) whose first term is \(a_{1}\) is found using the formula \(S_{n}=\frac{n}{2}\left(a_{1}+a_{n}\right)\)
5 step solution
Problem 4
True or False The notation \(a_{5}\) represents the fifth term of a sequence.
4 step solution
Problem 5
Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 2+5+8+\cdots+(3 n-1)=\frac{1}{2} n(3 n+1) $$
5 step solution
Problem 5
If a series does not converge, it is called a(n) _____ series. (a) arithmetic (b) divergent (c) geometric (d) recursive
4 step solution
Problem 5
True or False If \(n \geq 2\) is an integer, then $$ n !=n(n-1) \cdots 3 \cdot 2 \cdot 1 $$
4 step solution
Problem 6
Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 1+4+7+\cdots+(3 n-2)=\frac{1}{2} n(3 n-1) $$
6 step solution
Problem 6
Multiple Choice If \(a_{n}=-2 n+7\) is the \(n\) th term of an arithmetic sequence, the first term is _____. (a) -2 (b) 0 (c) 5 (d) 7
4 step solution
Problem 6
Multiple Choice The sequence \(a_{1}=5, a_{n}=3 a_{n-1}\) is an example of a(n) _______ sequence. (a) alternating (b) recursive (c) Fibonacci (d) summation
3 step solution
Problem 7
Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 1+2+2^{2}+\cdots+2^{n-1}=2^{n}-1 $$
4 step solution
Problem 7
In Problems 7 -16, show that each sequence is arithmetic. Find the common difference, and list the first four terms. $$ \left\\{s_{n}\right\\}=\\{n+4\\} $$
4 step solution
Problem 7
The notation $$ a_{1}+a_{2}+a_{3}+\cdots+a_{n}=\sum_{k=1}^{n} a_{k} $$ is an example of _______ notation.
4 step solution
Problem 8
Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 1+3+3^{2}+\cdots+3^{n-1}=\frac{1}{2}\left(3^{n}-1\right) $$
4 step solution
Problem 8
For a geometric sequence with first term \(a_{1}\) and common ratio \(r,\) where \(r \neq 0, r \neq 1,\) the sum of the first \(n\) terms is \(S_{n}=a_{1} \cdot \frac{1-r^{n}}{1-r}\)
5 step solution
Problem 8
Show that each sequence is arithmetic. Find the common difference, and list the first four terms. $$ \left\\{s_{n}\right\\}=\\{n-5\\} $$
3 step solution
Problem 8
Multiple Choice \(\sum_{k=1}^{n} k=1+2+3+\cdots+n=\) ________. (a) \(n !\) (b) \(\frac{n(n+1)}{2}\) (c) \(n k\) (d) \(\frac{n(n+1)(2 n+1)}{6}\)
3 step solution
Problem 9
Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 1+4+4^{2}+\cdots+4^{n-1}=\frac{1}{3}\left(4^{n}-1\right) $$
4 step solution
Problem 9
Show that each sequence is geometric. Then find the common ratio and list the first four terms. $$ \left\\{s_{n}\right\\}=\left\\{4^{n}\right\\} $$
4 step solution
Problem 9
Evaluate each factorial expression. \(10 !\)
4 step solution
Problem 9
Show that each sequence is arithmetic. Find the common difference, and list the first four terms. $$ \left\\{a_{n}\right\\}=\\{2 n-5\\} $$
5 step solution
Problem 10
Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 1+5+5^{2}+\cdots+5^{n-1}=\frac{1}{4}\left(5^{n}-1\right) $$
5 step solution
Problem 10
Show that each sequence is geometric. Then find the common ratio and list the first four terms. $$ \left.s_{n}\right\\}=\left\\{(-5)^{n}\right\\} $$
4 step solution
Problem 10
Evaluate each factorial expression. \(9 !\)
4 step solution
Problem 10
Show that each sequence is arithmetic. Find the common difference, and list the first four terms. $$ \left\\{b_{n}\right\\}=\\{3 n+1\\} $$
7 step solution
Problem 11
Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{n(n+1)} \equiv \frac{n}{n+1} $$
5 step solution
Problem 11
Show that each sequence is geometric. Then find the common ratio and list the first four terms. $$ \left\\{a_{n}\right\\}=\left\\{-3\left(\frac{1}{2}\right)^{n}\right\\} $$
4 step solution
Problem 11
Evaluate each factorial expression. \(\frac{9 !}{6 !}\)
4 step solution
Problem 11
Show that each sequence is arithmetic. Find the common difference, and list the first four terms. $$ \left\\{c_{n}\right\\}=\\{6-2 n\\} $$
4 step solution
Problem 12
Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ \frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\cdots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1} $$
5 step solution
Problem 12
Evaluate each factorial expression. \(\frac{12 !}{10 !}\)
4 step solution
Problem 12
Show that each sequence is arithmetic. Find the common difference, and list the first four terms. $$ \left\\{a_{n}\right\\}=\\{4-2 n\\} $$
4 step solution
Problem 13
Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 1^{2}+2^{2}+3^{2}+\cdots+n^{2}=\frac{1}{6} n(n+1)(2 n+1) $$
6 step solution
Problem 13
Show that each sequence is geometric. Then find the common ratio and list the first four terms. $$ \left\\{c_{n}\right\\}=\left\\{\frac{2^{n-1}}{4}\right\\} $$
4 step solution
Problem 13
Evaluate each factorial expression. \(\frac{4 ! 11 !}{7 !}\)
4 step solution
Problem 13
Show that each sequence is arithmetic. Find the common difference, and list the first four terms. $$ \left\\{t_{n}\right\\}=\left\\{\frac{1}{2}-\frac{1}{3} n\right\\} $$
6 step solution
Problem 14
Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 1^{3}+2^{3}+3^{3}+\cdots+n^{3}=\frac{1}{4} n^{2}(n+1)^{2} $$
7 step solution
Problem 14
Show that each sequence is geometric. Then find the common ratio and list the first four terms. $$ \left\\{d_{n}\right\\}=\left\\{\frac{3^{n}}{9}\right\\} $$
4 step solution
Problem 14
Evaluate each factorial expression. \(\frac{5 ! 8 !}{3 !}\)
4 step solution
Problem 15
Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 4+3+2+\cdots+(5-n)=\frac{1}{2} n(9-n) $$
5 step solution
Problem 15
Show that each sequence is geometric. Then find the common ratio and list the first four terms. $$ \left\\{e_{n}\right\\}=\left\\{7^{n / 4}\right\\} $$
5 step solution
Problem 15
List the first five terms of each sequence. \(\left\\{s_{n}\right\\}=\\{n\\}\)
7 step solution
Problem 15
Show that each sequence is arithmetic. Find the common difference, and list the first four terms. $$ \left\\{s_{n}\right\\}=\left\\{\ln 3^{n}\right\\} $$
4 step solution
Problem 16
Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ -2-3-4-\cdots-(n+1)=-\frac{1}{2} n(n+3) $$
5 step solution