Chapter 13, Problem 1CRT

a.

To determine

To find: the $7th$ and $20th$ term and the limit of the sequence is exists.

Answer to Problem 1CRT

The $7th$ and $20th$ terms are $\frac{7}{15}\text{ and }\frac{20}{41}$ and the limit of the sequence is $\underset{n\to \infty }{\lim }\frac{n}{2n+1}$.

Explanation of Solution

Given information:

The sequence is $\frac{1}{3},\frac{2}{5},\frac{3}{7},\frac{4}{9},\frac{5}{11},\mathrm{...}$.

Calculation: to find the $7th$ and $20th$ terms, first write the formula of $nth$ term,

  $a_n=\frac{n}{2n+1}$

Now, substitute $n=7\text{ and }n=20$ in the formula,

  $\begin{array}{l}{a}_n=\frac{n}{2n+1}\\ a_7=\frac{7}{2\left(7\right)+1}\text{ [substitute }n=7]\\ a_7=\frac{7}{15}\text{ [simplify]}\\ a_{20}=\frac{20}{2\left(20\right)+1}\text{ [substitute }n=20]\\ a_{20}=\frac{20}{41}\text{ [simplify]}\end{array}$

Thus, The $7th$ and $20th$ terms are $\frac{7}{15}\text{ and }\frac{20}{41}$.

The function defined in the interval $[1,\infty )$ , then the limit of the sequence is:

  $\underset{n\to \infty }{\lim }\frac{n}{2n+1}$

b.

To determine

To find: the $7th$ and $20th$ term and the limit of the sequence is exists.

Answer to Problem 1CRT

The $7th$ and $20th$ terms are $\frac{99}{340}\text{ and }\frac{801}{7984}$ and the limit of the sequence is $\underset{n\to \infty }{\lim }\frac{2n^2+1}{n^3-n+4}$.

Explanation of Solution

Given information:

The sequence is $a_n=\frac{2n^2+1}{n^3-n+4}$.

Calculation:

Now, substitute $n=7\text{ and }n=20$ in the formula,

  $\begin{array}{l}{a}_n=\frac{2n^2+1}{n^3-n+4}\\ a_7=\frac{2{\left(7\right)}^2+1}{{\left(7\right)}^3-7+4}\text{ [substitute }n=7]\\ a_7=\frac{99}{340}\text{ [simplify]}\\ a_{20}=\frac{2{\left(20\right)}^2+1}{{\left(20\right)}^3-20+4}\text{ [substitute }n=20]\\ a_{20}=\frac{801}{7984}\text{ [simplify]}\end{array}$

Thus, The $7th$ and $20th$ terms are $\frac{99}{340}\text{ and }\frac{801}{7984}$.

The function defined in the interval $[1,\infty )$ , then the limit of the sequence is:

  $\underset{n\to \infty }{\lim }\frac{2n^2+1}{n^3-n+4}$

c.

To determine

To find: the $7th$ and $20th$ term and the limit of the sequence is exists.

Answer to Problem 1CRT

The $7th$ and $20th$ terms are $\frac{37}{2}\text{ and }\frac{115}{2}$ and the limit of the sequence is $\underset{n\to \infty }{\lim }a+\left(n-1\right)d$.

Explanation of Solution

Given information:

The arithmetic sequence, initial term $a=\frac{1}{2}$ and common difference is $d=3$.

Calculation: the $nth$ term formula for arithmetic sequence,

  $a_n=a+\left(n-1\right)d$

Now, substitute $n=7\text{ and }n=20$ in the formula,

  $\begin{array}{l}{a}_n=a+\left(n-1\right)d\\ a_7=\frac{1}{2}+\left(7-1\right)\left(3\right)\text{ [substitute }n=7]\\ a_7=\frac{37}{2}\text{ [simplify]}\\ a_7=\frac{1}{2}+\left(20-1\right)\left(3\right)\text{ [substitute }n=20]\\ a_{20}=\frac{115}{2}\text{ [simplify]}\end{array}$

Thus, The $7th$ and $20th$ terms are $\frac{37}{2}\text{ and }\frac{115}{2}$.

The function defined in the interval $[1,\infty )$ , then the limit of the sequence is:

  $\underset{n\to \infty }{\lim }a+\left(n-1\right)d$

d.

To determine

To find: the $7th$ and $20th$ term and the limit of the sequence is exists.

Answer to Problem 1CRT

The $7th$ and $20th$ terms are $\frac{31250}{7776}\text{ and }\frac{2{\left(5\right)}^{19}}{{\left(6\right)}^{18}}$ and the limit of the sequence is $\underset{n\to \infty }{\lim }a{r}^{n-1}$.

Explanation of Solution

Given information:

The geometric sequence, initial term $a=12$ and common ratio is $r=\frac{5}{6}$.

Calculation: the $nth$ term formula for geometric sequence,

  $a_n=a{r}^{n-1}$

Now, substitute $n=7\text{ and }n=20$ in the formula,

  $\begin{array}{l}{a}_n=a{r}^{n-1}\\ a_7=12{\left(\frac{5}{6}\right)}^{7-1}\text{ [substitute }n=7]\\ a_7=\frac{31250}{7776}\text{ [simplify]}\\ a_7=12{\left(\frac{5}{6}\right)}^{20-1}\text{ [substitute }n=20]\\ a_{20}=\frac{2{\left(5\right)}^{19}}{{\left(6\right)}^{18}}\text{ [simplify]}\end{array}$

Thus, The $7th$ and $20th$ terms are $\frac{31250}{7776}\text{ and }\frac{2{\left(5\right)}^{19}}{{\left(6\right)}^{18}}$.

The function defined in the interval $[1,\infty )$ , then the limit of the sequence is:

  $\underset{n\to \infty }{\lim }a{r}^{n-1}$

e.

To determine

To find: the $7th$ and $20th$ term and the limit of the sequence is exists.

Answer to Problem 1CRT

The $7th$ and $20th$ terms are $\text{0}\text{.64 and }-\text{5242}\text{.88}$ and the limit of the sequence is $\underset{n\to \infty }{\lim }-2a_{n-1}$.

Explanation of Solution

Given information:

The sequence, initial term $a_1=0.01$ and $a_n=-2a_{n-1}$.

Calculation: the $nth$ term formula for geometric sequence,

  $a_n=-2a_{n-1}$

Now, first substitute and then simplify,

Thus, The $7th$ and $20th$ terms are $\text{0}\text{.64 and }-\text{5242}\text{.88}$.

The function defined in the interval $[1,\infty )$ , then the limit of the sequence is:

  $\underset{n\to \infty }{\lim }-2a_{n-1}$