Chapter 11.6, Problem 1ST

(a)

To determine

To find: That if the given series is arithmetic, geometric series or neither and rewrite the series using sigma notation.

Answer to Problem 1ST

The series is geometric series and the sigma notation is ${\displaystyle {\sum }_{n=1}^{4}a{r}^{n-1}=}{\displaystyle {\sum }_{n=1}^{4}27{(-\frac{2}{3})}^{n-1}}$ .

Explanation of Solution

Given Information:

The given series is $27-18+12-8$ .

Every term that is given is $\frac{-2}{3}$ times that of the previous term therefore it has a common ratio and is geometric series where $r=\frac{-2}{3}$ .

In the given series the first term $a=27$

Therefore, in sigma notation it is ${\displaystyle {\sum }_{n=1}^{4}a{r}^{n-1}=}{\displaystyle {\sum }_{n=1}^{4}27{(-\frac{2}{3})}^{n-1}}$

Hence, the series is geometric series.

(b)

To determine

To find: That if the given series is arithmetic, geometric series or neither and rewrite the series using sigma notation.

Answer to Problem 1ST

The series is neither arithmetic nor geometric and the sigma notation is ${\displaystyle {\sum }_{n=1}^{15}\frac{n}{n+1}}$ .

Explanation of Solution

Given Information:

The given series is $\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\mathrm{..........}+\frac{15}{16}$ .

  $\frac{\frac{2}{3}}{\frac{1}{2}}=\frac{4}{3}$ and $\frac{\frac{3}{4}}{\frac{2}{3}}=\frac{9}{8}$

Again, $\frac{2}{3}-\frac{1}{2}=\frac{4-3}{6}=\frac{1}{6}$ and $\frac{3}{4}-\frac{2}{3}=\frac{9-8}{12}=\frac{1}{12}$

So, the index of summation $n$ is the numerator and $n+1$ is the denominator.

Therefore, in sigma notation it is ${\displaystyle {\sum }_{n=1}^{15}\frac{n}{n+1}}$ .

Hence, the series is neither geometric nor arithmetic.

(c)

To determine

To find: That if the given series is arithmetic, geometric series or neither and rewrite the series using sigma notation.

Answer to Problem 1ST

The series is arithmetic series and the sigma notation is ${\displaystyle {\sum }_{n=1}^7(23-3n)}$ .

Explanation of Solution

Given Information:

The given series is $20+17+14+\mathrm{...}+2$ .

If we try to find the n of the last term then,

  $\begin{array}{c}20-3n=2\\ -3n=-18\\ n=6\end{array}$

Therefore, ${\displaystyle {\sum }_{n=0}^6(20-3n)}$ is an arithmetic series.

Thus, in sigma notation it is ${\displaystyle {\sum }_{n=1}^7(20-3(n-1))=}$

  ${\displaystyle {\sum }_{n=1}^7(23-3n)}$ .

Hence, the series is arithmetic series.

(d)

To determine

To find: That if the given series is arithmetic, geometric series or neither and rewrite the series using sigma notation.

Answer to Problem 1ST

The series is neither arithmetic nor geometric and sigma notation is ${\displaystyle {\sum }_{n=1}^4{n}^2}$ .

Explanation of Solution

Given Information:

The given series is $1+4+9+16+\mathrm{....}$ .

The given sequence $1+4+9+16+\mathrm{....}$ is neither arithmetic nor geometric as there is no common multiplier. Though it has a sigma notation of ${\displaystyle {\sum }_{n=1}^4{n}^2}$ .

Hence, the series is neither arithmetic nor geometric.