Chapter 9.4, Problem 50PPSE

Solution

a.

To write two explicit formulas for arithmetic sequences.

The two explicit formulas for sequences are $a_n=n+2$ and $a_n=2n+1$ .

Consider the sequences $3,4,5,6,\mathrm{...}$ and $3,5,7,9,\mathrm{...}$ .

The nth term of $3,4,5,6,\mathrm{...}$ is $a_n=n+2$ , and the nth term of $3,5,7,9,\mathrm{...}$ is $a_n=2n+1$ .

Therefore, the two explicit formulas for sequences are $a_n=n+2$ and $a_n=2n+1$ .

b.

To write the first five terms of each related series.

The first sequence is $3,4,5,6,\mathrm{...}$ .

The first five terms of the related series for the sequence $3,4,5,6,\mathrm{...}$ can be written as:

  $3+4+5+6+7$

Similarly, the first five terms of the related series of the sequence $3,5,7,9,\mathrm{...}$ can be written as:

  $3+5+7+9+11$

c.

To rewrite the series using summation notation.

The series with summation notations are ${\displaystyle \sum _{n=1}^5\left(n+2\right)}$ and ${\displaystyle \sum _{n=1}^5\left(2n+1\right)}$ .

The two series from part (b) are $3+4+5+6+7$ and $3+5+7+9+11$ .

The nth term of $3+4+5+6+7$ is $a_n=n+2$ , and it has 5 terms. So, the series can be rewritten using summation notation as ${\displaystyle \sum _{n=1}^5\left(n+2\right)}$ .

The nth term of $3+5+7+9+11$ is $a_n=2n+1$ , and it has 5 terms. So, the series can be rewritten using summation notation as ${\displaystyle \sum _{n=1}^5\left(2n+1\right)}$ .

d.

To evaluate each series.

The two series are ${\displaystyle \sum _{n=1}^5\left(n+2\right)}$ and ${\displaystyle \sum _{n=1}^5\left(2n+1\right)}$ .

Now, evaluate the series ${\displaystyle \sum _{n=1}^5\left(n+2\right)}$ .

  $\begin{array}{c}{\displaystyle \sum _{n=1}^5\left(n+2\right)}={\displaystyle \sum _{n=1}^{5}n}+{\displaystyle \sum _{n=1}^{5}2}\\ =\frac{5\left(5+1\right)}{2}+2\cdot 5\\ =15+10\\ =25\end{array}$

Now, evaluate the series ${\displaystyle \sum _{n=1}^5\left(2n+1\right)}$ .

  $\begin{array}{c}{\displaystyle \sum _{n=1}^5\left(2n+1\right)}=2{\displaystyle \sum _{n=1}^{5}n}+{\displaystyle \sum _{n=1}^{5}1}\\ =2\cdot \frac{5\left(5+1\right)}{2}+1\cdot 5\\ =2\cdot 15+5\\ =35\end{array}$