a) The first three terms of the sequence having the n th term is given by a n 2 .

Question

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Answer 1

Question

Chapter 11, Problem 1CRE

To determine

a)

The first three terms of the sequence having the n^th term is given by $a_{n}^{2}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of $a_{n}^{2}$ is given by

$\begin{array}{l} a_{1}^{2} = \frac{4}{1!} = 3 \\ a_{2}^{2} = \frac{1}{4} \\ a_{3}^{2} = 0 \end{array}$

Explanation of Solution

Given:

$a_{n} = \frac{n - 3}{n!}$

Calculation:

Here, we have

$a_{n} = \frac{n - 3}{n!} \Rightarrow a_{n}^{2} = {(\frac{n - 3}{n!})}^{2}$

The first three terms of $a_{n}^{2}$ is given by

$\begin{array}{l} a_{1}^{2} = {(\frac{1 - 3}{1!})}^{2} = \frac{4}{1!} \\ a_{2}^{2} = {(\frac{2 - 3}{2!})}^{2} = \frac{1}{(2!) (2!)} = \frac{1}{4} \\ a_{3}^{2} = {(\frac{3 - 3}{3!})}^{2} = 0 \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} a_{1}^{2} = \frac{4}{1!} = 3 \\ a_{2}^{2} = \frac{1}{4} \\ a_{3}^{2} = 0 \end{array}$

To determine

b)

The first three terms of the sequence having the n^th term is given by $b_{n}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of the sequence is given by

$\begin{array}{l} b_{1} = \frac{1}{4!} \\ b_{2} = \frac{2}{5!} \\ b_{3} = \frac{3}{6!} \end{array}$

Explanation of Solution

Given:

$b_{n} = a_{n + 3}$ , where $a_{n} = \frac{n - 3}{n!}$

Calculation:

Here, we have

$a_{n} = \frac{n - 3}{n!} \Rightarrow b_{n} = a_{n + 3} = (\frac{n + 3 - 3}{(n + 3)!}) = \frac{n}{(n + 3)!}$

The first three terms of $b_{n} = a_{n + 3}$ is given by

$\begin{array}{l} b_{1} = \frac{1}{(1 + 3)!} = \frac{1}{4!} \\ b_{2} = \frac{2}{(2 + 3)!} = \frac{2}{5!} \\ b_{3} = \frac{3}{(3 + 3)!} = \frac{3}{6!} \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} b_{1} = \frac{1}{4!} \\ b_{2} = \frac{2}{5!} \\ b_{3} = \frac{3}{6!} \end{array}$

To determine

c)

The first three terms of the sequence having the n^th term is given by $a_{n} b_{n}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = \frac{- 1}{12} \\ t_{2} = \frac{- 1}{120} \\ t_{3} = 0 \end{array}$

Explanation of Solution

Given:

$t_{n} = a_{n} b_{n}$ , where $a_{n} = \frac{n - 3}{n!}$ and $b_{n} = a_{n + 3}$

Calculation:

Here, we have

$a_{n} = \frac{n - 3}{n!} \Rightarrow t_{n} = a_{n} b_{n} = (\frac{n - 3}{n!}) (\frac{n + 3 - 3}{(n + 3)!}) = \frac{n - 3}{n!} \frac{n}{(n + 3)!}$

The first three terms of $t_{n} = a_{n} b_{n} = \frac{n - 3}{n!} \frac{n}{(n + 3)!}$ is given by

$\begin{array}{l} t_{1} = \frac{1 - 3}{1!} \frac{1}{(1 + 3)!} = \frac{- 2}{4!} = \frac{- 2}{24} = \frac{- 1}{12} \\ t_{2} = \frac{2 - 3}{2!} \frac{2}{(2 + 3)!} = \frac{- 2}{(2!) (5!)} = \frac{- 2}{240} = \frac{- 1}{120} \\ t_{3} = 0 \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = \frac{- 1}{12} \\ t_{2} = \frac{- 1}{120} \\ t_{3} = 0 \end{array}$

To determine

d)

The first three terms of the sequence having the n^th term is given by $2 a_{n + 1} - 3 a_{n}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = 5 \\ t_{2} = \frac{3}{2} \\ t_{3} = \frac{1}{12} \end{array}$

Explanation of Solution

Given:

$t_{n} = 2 a_{n + 1} - 3 a_{n}$ , where $a_{n} = \frac{n - 3}{n!}$

Calculation:

Here, we have

$\begin{array}{l} a_{n} = \frac{n - 3}{n!} \\ \Rightarrow t_{n} = 2 a_{n + 1} - 3 a_{n} \\ = 2 (\frac{n + 1 - 3}{(n + 1)!}) - 3 (\frac{n - 3}{n!}) \\ = 2 (\frac{n - 2}{(n + 1) n!}) - 3 (\frac{n - 3}{n!}) \\ = \frac{(2 n - 4) - 3 (n - 3) (n + 1)}{(n + 1) n!} \\ = \frac{(2 n - 4) - 3 (n^{2} - 2 n - 3)}{(n + 1)!} \\ ∴ t_{n} = 2 a_{n + 1} - 3 a_{n} = \frac{- 3 n^{2} + 8 n + 5}{(n + 1)!} \end{array}$

The first three terms of $t_{n} = 2 a_{n + 1} - 3 a_{n} = \frac{- 3 n^{2} + 8 n + 5}{(n + 1)!}$ is given by

$\begin{array}{l} t_{1} = \frac{- 3 {(1)}^{2} + 8 (1) + 5}{(1 + 1)!} = \frac{10}{2!} = 5 \\ t_{2} = \frac{- 3 {(2)}^{2} + 8 (2) + 5}{(2 + 1)!} = \frac{9}{3!} = \frac{3}{2} \\ t_{3} = \frac{- 3 {(3)}^{2} + 8 (3) + 5}{(3 + 1)!} = \frac{2}{4!} = \frac{1}{12} \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = 5 \\ t_{2} = \frac{3}{2} \\ t_{3} = \frac{1}{12} \end{array}$

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17. Suppose we know that the graph below is the graph of a solution to dy/dt = f(t). (a) How much of the slope field can you sketch from this information? [Hint: Note that the differential equation depends only on t.] (b) What can you say about the solu- tion with y(0) = 2? (For example, can you sketch the graph of this so- lution?) y(0) = 1 y AN

(b) Find the (instantaneous) rate of change of y at x = 5. In the previous part, we found the average rate of change for several intervals of decreasing size starting at x = 5. The instantaneous rate of change of fat x = 5 is the limit of the average rate of change over the interval [x, x + h] as h approaches 0. This is given by the derivative in the following limit. lim h→0 - f(x + h) − f(x) h The first step to find this limit is to compute f(x + h). Recall that this means replacing the input variable x with the expression x + h in the rule defining f. f(x + h) = (x + h)² - 5(x+ h) = 2xh+h2_ x² + 2xh + h² 5✔ - 5 )x - 5h Step 4 - The second step for finding the derivative of fat x is to find the difference f(x + h) − f(x). - f(x + h) f(x) = = (x² x² + 2xh + h² - ])- = 2x + h² - 5h ])x-5h) - (x² - 5x) = ]) (2x + h - 5) Macbook Pro

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Answer 2

Question

Chapter 11, Problem 1CRE

To determine

a)

The first three terms of the sequence having the n^th term is given by $a_{n}^{2}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of $a_{n}^{2}$ is given by

$\begin{array}{l} a_{1}^{2} = \frac{4}{1!} = 3 \\ a_{2}^{2} = \frac{1}{4} \\ a_{3}^{2} = 0 \end{array}$

Explanation of Solution

Given:

$a_{n} = \frac{n - 3}{n!}$

Calculation:

Here, we have

$a_{n} = \frac{n - 3}{n!} \Rightarrow a_{n}^{2} = {(\frac{n - 3}{n!})}^{2}$

The first three terms of $a_{n}^{2}$ is given by

$\begin{array}{l} a_{1}^{2} = {(\frac{1 - 3}{1!})}^{2} = \frac{4}{1!} \\ a_{2}^{2} = {(\frac{2 - 3}{2!})}^{2} = \frac{1}{(2!) (2!)} = \frac{1}{4} \\ a_{3}^{2} = {(\frac{3 - 3}{3!})}^{2} = 0 \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} a_{1}^{2} = \frac{4}{1!} = 3 \\ a_{2}^{2} = \frac{1}{4} \\ a_{3}^{2} = 0 \end{array}$

To determine

b)

The first three terms of the sequence having the n^th term is given by $b_{n}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of the sequence is given by

$\begin{array}{l} b_{1} = \frac{1}{4!} \\ b_{2} = \frac{2}{5!} \\ b_{3} = \frac{3}{6!} \end{array}$

Explanation of Solution

Given:

$b_{n} = a_{n + 3}$ , where $a_{n} = \frac{n - 3}{n!}$

Calculation:

Here, we have

$a_{n} = \frac{n - 3}{n!} \Rightarrow b_{n} = a_{n + 3} = (\frac{n + 3 - 3}{(n + 3)!}) = \frac{n}{(n + 3)!}$

The first three terms of $b_{n} = a_{n + 3}$ is given by

$\begin{array}{l} b_{1} = \frac{1}{(1 + 3)!} = \frac{1}{4!} \\ b_{2} = \frac{2}{(2 + 3)!} = \frac{2}{5!} \\ b_{3} = \frac{3}{(3 + 3)!} = \frac{3}{6!} \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} b_{1} = \frac{1}{4!} \\ b_{2} = \frac{2}{5!} \\ b_{3} = \frac{3}{6!} \end{array}$

To determine

c)

The first three terms of the sequence having the n^th term is given by $a_{n} b_{n}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = \frac{- 1}{12} \\ t_{2} = \frac{- 1}{120} \\ t_{3} = 0 \end{array}$

Explanation of Solution

Given:

$t_{n} = a_{n} b_{n}$ , where $a_{n} = \frac{n - 3}{n!}$ and $b_{n} = a_{n + 3}$

Calculation:

Here, we have

$a_{n} = \frac{n - 3}{n!} \Rightarrow t_{n} = a_{n} b_{n} = (\frac{n - 3}{n!}) (\frac{n + 3 - 3}{(n + 3)!}) = \frac{n - 3}{n!} \frac{n}{(n + 3)!}$

The first three terms of $t_{n} = a_{n} b_{n} = \frac{n - 3}{n!} \frac{n}{(n + 3)!}$ is given by

$\begin{array}{l} t_{1} = \frac{1 - 3}{1!} \frac{1}{(1 + 3)!} = \frac{- 2}{4!} = \frac{- 2}{24} = \frac{- 1}{12} \\ t_{2} = \frac{2 - 3}{2!} \frac{2}{(2 + 3)!} = \frac{- 2}{(2!) (5!)} = \frac{- 2}{240} = \frac{- 1}{120} \\ t_{3} = 0 \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = \frac{- 1}{12} \\ t_{2} = \frac{- 1}{120} \\ t_{3} = 0 \end{array}$

To determine

d)

The first three terms of the sequence having the n^th term is given by $2 a_{n + 1} - 3 a_{n}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = 5 \\ t_{2} = \frac{3}{2} \\ t_{3} = \frac{1}{12} \end{array}$

Explanation of Solution

Given:

$t_{n} = 2 a_{n + 1} - 3 a_{n}$ , where $a_{n} = \frac{n - 3}{n!}$

Calculation:

Here, we have

$\begin{array}{l} a_{n} = \frac{n - 3}{n!} \\ \Rightarrow t_{n} = 2 a_{n + 1} - 3 a_{n} \\ = 2 (\frac{n + 1 - 3}{(n + 1)!}) - 3 (\frac{n - 3}{n!}) \\ = 2 (\frac{n - 2}{(n + 1) n!}) - 3 (\frac{n - 3}{n!}) \\ = \frac{(2 n - 4) - 3 (n - 3) (n + 1)}{(n + 1) n!} \\ = \frac{(2 n - 4) - 3 (n^{2} - 2 n - 3)}{(n + 1)!} \\ ∴ t_{n} = 2 a_{n + 1} - 3 a_{n} = \frac{- 3 n^{2} + 8 n + 5}{(n + 1)!} \end{array}$

The first three terms of $t_{n} = 2 a_{n + 1} - 3 a_{n} = \frac{- 3 n^{2} + 8 n + 5}{(n + 1)!}$ is given by

$\begin{array}{l} t_{1} = \frac{- 3 {(1)}^{2} + 8 (1) + 5}{(1 + 1)!} = \frac{10}{2!} = 5 \\ t_{2} = \frac{- 3 {(2)}^{2} + 8 (2) + 5}{(2 + 1)!} = \frac{9}{3!} = \frac{3}{2} \\ t_{3} = \frac{- 3 {(3)}^{2} + 8 (3) + 5}{(3 + 1)!} = \frac{2}{4!} = \frac{1}{12} \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = 5 \\ t_{2} = \frac{3}{2} \\ t_{3} = \frac{1}{12} \end{array}$

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Answer 3

Question

Chapter 11, Problem 1CRE

To determine

a)

The first three terms of the sequence having the n^th term is given by $a_{n}^{2}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of $a_{n}^{2}$ is given by

$\begin{array}{l} a_{1}^{2} = \frac{4}{1!} = 3 \\ a_{2}^{2} = \frac{1}{4} \\ a_{3}^{2} = 0 \end{array}$

Explanation of Solution

Given:

$a_{n} = \frac{n - 3}{n!}$

Calculation:

Here, we have

$a_{n} = \frac{n - 3}{n!} \Rightarrow a_{n}^{2} = {(\frac{n - 3}{n!})}^{2}$

The first three terms of $a_{n}^{2}$ is given by

$\begin{array}{l} a_{1}^{2} = {(\frac{1 - 3}{1!})}^{2} = \frac{4}{1!} \\ a_{2}^{2} = {(\frac{2 - 3}{2!})}^{2} = \frac{1}{(2!) (2!)} = \frac{1}{4} \\ a_{3}^{2} = {(\frac{3 - 3}{3!})}^{2} = 0 \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} a_{1}^{2} = \frac{4}{1!} = 3 \\ a_{2}^{2} = \frac{1}{4} \\ a_{3}^{2} = 0 \end{array}$

To determine

b)

The first three terms of the sequence having the n^th term is given by $b_{n}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of the sequence is given by

$\begin{array}{l} b_{1} = \frac{1}{4!} \\ b_{2} = \frac{2}{5!} \\ b_{3} = \frac{3}{6!} \end{array}$

Explanation of Solution

Given:

$b_{n} = a_{n + 3}$ , where $a_{n} = \frac{n - 3}{n!}$

Calculation:

Here, we have

$a_{n} = \frac{n - 3}{n!} \Rightarrow b_{n} = a_{n + 3} = (\frac{n + 3 - 3}{(n + 3)!}) = \frac{n}{(n + 3)!}$

The first three terms of $b_{n} = a_{n + 3}$ is given by

$\begin{array}{l} b_{1} = \frac{1}{(1 + 3)!} = \frac{1}{4!} \\ b_{2} = \frac{2}{(2 + 3)!} = \frac{2}{5!} \\ b_{3} = \frac{3}{(3 + 3)!} = \frac{3}{6!} \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} b_{1} = \frac{1}{4!} \\ b_{2} = \frac{2}{5!} \\ b_{3} = \frac{3}{6!} \end{array}$

To determine

c)

The first three terms of the sequence having the n^th term is given by $a_{n} b_{n}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = \frac{- 1}{12} \\ t_{2} = \frac{- 1}{120} \\ t_{3} = 0 \end{array}$

Explanation of Solution

Given:

$t_{n} = a_{n} b_{n}$ , where $a_{n} = \frac{n - 3}{n!}$ and $b_{n} = a_{n + 3}$

Calculation:

Here, we have

$a_{n} = \frac{n - 3}{n!} \Rightarrow t_{n} = a_{n} b_{n} = (\frac{n - 3}{n!}) (\frac{n + 3 - 3}{(n + 3)!}) = \frac{n - 3}{n!} \frac{n}{(n + 3)!}$

The first three terms of $t_{n} = a_{n} b_{n} = \frac{n - 3}{n!} \frac{n}{(n + 3)!}$ is given by

$\begin{array}{l} t_{1} = \frac{1 - 3}{1!} \frac{1}{(1 + 3)!} = \frac{- 2}{4!} = \frac{- 2}{24} = \frac{- 1}{12} \\ t_{2} = \frac{2 - 3}{2!} \frac{2}{(2 + 3)!} = \frac{- 2}{(2!) (5!)} = \frac{- 2}{240} = \frac{- 1}{120} \\ t_{3} = 0 \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = \frac{- 1}{12} \\ t_{2} = \frac{- 1}{120} \\ t_{3} = 0 \end{array}$

To determine

d)

The first three terms of the sequence having the n^th term is given by $2 a_{n + 1} - 3 a_{n}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = 5 \\ t_{2} = \frac{3}{2} \\ t_{3} = \frac{1}{12} \end{array}$

Explanation of Solution

Given:

$t_{n} = 2 a_{n + 1} - 3 a_{n}$ , where $a_{n} = \frac{n - 3}{n!}$

Calculation:

Here, we have

$\begin{array}{l} a_{n} = \frac{n - 3}{n!} \\ \Rightarrow t_{n} = 2 a_{n + 1} - 3 a_{n} \\ = 2 (\frac{n + 1 - 3}{(n + 1)!}) - 3 (\frac{n - 3}{n!}) \\ = 2 (\frac{n - 2}{(n + 1) n!}) - 3 (\frac{n - 3}{n!}) \\ = \frac{(2 n - 4) - 3 (n - 3) (n + 1)}{(n + 1) n!} \\ = \frac{(2 n - 4) - 3 (n^{2} - 2 n - 3)}{(n + 1)!} \\ ∴ t_{n} = 2 a_{n + 1} - 3 a_{n} = \frac{- 3 n^{2} + 8 n + 5}{(n + 1)!} \end{array}$

The first three terms of $t_{n} = 2 a_{n + 1} - 3 a_{n} = \frac{- 3 n^{2} + 8 n + 5}{(n + 1)!}$ is given by

$\begin{array}{l} t_{1} = \frac{- 3 {(1)}^{2} + 8 (1) + 5}{(1 + 1)!} = \frac{10}{2!} = 5 \\ t_{2} = \frac{- 3 {(2)}^{2} + 8 (2) + 5}{(2 + 1)!} = \frac{9}{3!} = \frac{3}{2} \\ t_{3} = \frac{- 3 {(3)}^{2} + 8 (3) + 5}{(3 + 1)!} = \frac{2}{4!} = \frac{1}{12} \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = 5 \\ t_{2} = \frac{3}{2} \\ t_{3} = \frac{1}{12} \end{array}$

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Answer 4

Question

Chapter 11, Problem 1CRE

To determine

a)

The first three terms of the sequence having the n^th term is given by $a_{n}^{2}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of $a_{n}^{2}$ is given by

$\begin{array}{l} a_{1}^{2} = \frac{4}{1!} = 3 \\ a_{2}^{2} = \frac{1}{4} \\ a_{3}^{2} = 0 \end{array}$

Explanation of Solution

Given:

$a_{n} = \frac{n - 3}{n!}$

Calculation:

Here, we have

$a_{n} = \frac{n - 3}{n!} \Rightarrow a_{n}^{2} = {(\frac{n - 3}{n!})}^{2}$

The first three terms of $a_{n}^{2}$ is given by

$\begin{array}{l} a_{1}^{2} = {(\frac{1 - 3}{1!})}^{2} = \frac{4}{1!} \\ a_{2}^{2} = {(\frac{2 - 3}{2!})}^{2} = \frac{1}{(2!) (2!)} = \frac{1}{4} \\ a_{3}^{2} = {(\frac{3 - 3}{3!})}^{2} = 0 \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} a_{1}^{2} = \frac{4}{1!} = 3 \\ a_{2}^{2} = \frac{1}{4} \\ a_{3}^{2} = 0 \end{array}$

To determine

b)

The first three terms of the sequence having the n^th term is given by $b_{n}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of the sequence is given by

$\begin{array}{l} b_{1} = \frac{1}{4!} \\ b_{2} = \frac{2}{5!} \\ b_{3} = \frac{3}{6!} \end{array}$

Explanation of Solution

Given:

$b_{n} = a_{n + 3}$ , where $a_{n} = \frac{n - 3}{n!}$

Calculation:

Here, we have

$a_{n} = \frac{n - 3}{n!} \Rightarrow b_{n} = a_{n + 3} = (\frac{n + 3 - 3}{(n + 3)!}) = \frac{n}{(n + 3)!}$

The first three terms of $b_{n} = a_{n + 3}$ is given by

$\begin{array}{l} b_{1} = \frac{1}{(1 + 3)!} = \frac{1}{4!} \\ b_{2} = \frac{2}{(2 + 3)!} = \frac{2}{5!} \\ b_{3} = \frac{3}{(3 + 3)!} = \frac{3}{6!} \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} b_{1} = \frac{1}{4!} \\ b_{2} = \frac{2}{5!} \\ b_{3} = \frac{3}{6!} \end{array}$

To determine

c)

The first three terms of the sequence having the n^th term is given by $a_{n} b_{n}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = \frac{- 1}{12} \\ t_{2} = \frac{- 1}{120} \\ t_{3} = 0 \end{array}$

Explanation of Solution

Given:

$t_{n} = a_{n} b_{n}$ , where $a_{n} = \frac{n - 3}{n!}$ and $b_{n} = a_{n + 3}$

Calculation:

Here, we have

$a_{n} = \frac{n - 3}{n!} \Rightarrow t_{n} = a_{n} b_{n} = (\frac{n - 3}{n!}) (\frac{n + 3 - 3}{(n + 3)!}) = \frac{n - 3}{n!} \frac{n}{(n + 3)!}$

The first three terms of $t_{n} = a_{n} b_{n} = \frac{n - 3}{n!} \frac{n}{(n + 3)!}$ is given by

$\begin{array}{l} t_{1} = \frac{1 - 3}{1!} \frac{1}{(1 + 3)!} = \frac{- 2}{4!} = \frac{- 2}{24} = \frac{- 1}{12} \\ t_{2} = \frac{2 - 3}{2!} \frac{2}{(2 + 3)!} = \frac{- 2}{(2!) (5!)} = \frac{- 2}{240} = \frac{- 1}{120} \\ t_{3} = 0 \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = \frac{- 1}{12} \\ t_{2} = \frac{- 1}{120} \\ t_{3} = 0 \end{array}$

To determine

d)

The first three terms of the sequence having the n^th term is given by $2 a_{n + 1} - 3 a_{n}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = 5 \\ t_{2} = \frac{3}{2} \\ t_{3} = \frac{1}{12} \end{array}$

Explanation of Solution

Given:

$t_{n} = 2 a_{n + 1} - 3 a_{n}$ , where $a_{n} = \frac{n - 3}{n!}$

Calculation:

Here, we have

$\begin{array}{l} a_{n} = \frac{n - 3}{n!} \\ \Rightarrow t_{n} = 2 a_{n + 1} - 3 a_{n} \\ = 2 (\frac{n + 1 - 3}{(n + 1)!}) - 3 (\frac{n - 3}{n!}) \\ = 2 (\frac{n - 2}{(n + 1) n!}) - 3 (\frac{n - 3}{n!}) \\ = \frac{(2 n - 4) - 3 (n - 3) (n + 1)}{(n + 1) n!} \\ = \frac{(2 n - 4) - 3 (n^{2} - 2 n - 3)}{(n + 1)!} \\ ∴ t_{n} = 2 a_{n + 1} - 3 a_{n} = \frac{- 3 n^{2} + 8 n + 5}{(n + 1)!} \end{array}$

The first three terms of $t_{n} = 2 a_{n + 1} - 3 a_{n} = \frac{- 3 n^{2} + 8 n + 5}{(n + 1)!}$ is given by

$\begin{array}{l} t_{1} = \frac{- 3 {(1)}^{2} + 8 (1) + 5}{(1 + 1)!} = \frac{10}{2!} = 5 \\ t_{2} = \frac{- 3 {(2)}^{2} + 8 (2) + 5}{(2 + 1)!} = \frac{9}{3!} = \frac{3}{2} \\ t_{3} = \frac{- 3 {(3)}^{2} + 8 (3) + 5}{(3 + 1)!} = \frac{2}{4!} = \frac{1}{12} \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = 5 \\ t_{2} = \frac{3}{2} \\ t_{3} = \frac{1}{12} \end{array}$

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Answer 5

Chapter 11, Problem 1CRE

To determine

a)

The first three terms of the sequence having the n^th term is given by $a_{n}^{2}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of $a_{n}^{2}$ is given by

$\begin{array}{l} a_{1}^{2} = \frac{4}{1!} = 3 \\ a_{2}^{2} = \frac{1}{4} \\ a_{3}^{2} = 0 \end{array}$

Explanation of Solution

Given:

$a_{n} = \frac{n - 3}{n!}$

Calculation:

Here, we have

$a_{n} = \frac{n - 3}{n!} \Rightarrow a_{n}^{2} = {(\frac{n - 3}{n!})}^{2}$

The first three terms of $a_{n}^{2}$ is given by

$\begin{array}{l} a_{1}^{2} = {(\frac{1 - 3}{1!})}^{2} = \frac{4}{1!} \\ a_{2}^{2} = {(\frac{2 - 3}{2!})}^{2} = \frac{1}{(2!) (2!)} = \frac{1}{4} \\ a_{3}^{2} = {(\frac{3 - 3}{3!})}^{2} = 0 \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} a_{1}^{2} = \frac{4}{1!} = 3 \\ a_{2}^{2} = \frac{1}{4} \\ a_{3}^{2} = 0 \end{array}$

To determine

b)

The first three terms of the sequence having the n^th term is given by $b_{n}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of the sequence is given by

$\begin{array}{l} b_{1} = \frac{1}{4!} \\ b_{2} = \frac{2}{5!} \\ b_{3} = \frac{3}{6!} \end{array}$

Explanation of Solution

Given:

$b_{n} = a_{n + 3}$ , where $a_{n} = \frac{n - 3}{n!}$

Calculation:

Here, we have

$a_{n} = \frac{n - 3}{n!} \Rightarrow b_{n} = a_{n + 3} = (\frac{n + 3 - 3}{(n + 3)!}) = \frac{n}{(n + 3)!}$

The first three terms of $b_{n} = a_{n + 3}$ is given by

$\begin{array}{l} b_{1} = \frac{1}{(1 + 3)!} = \frac{1}{4!} \\ b_{2} = \frac{2}{(2 + 3)!} = \frac{2}{5!} \\ b_{3} = \frac{3}{(3 + 3)!} = \frac{3}{6!} \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} b_{1} = \frac{1}{4!} \\ b_{2} = \frac{2}{5!} \\ b_{3} = \frac{3}{6!} \end{array}$

To determine

c)

The first three terms of the sequence having the n^th term is given by $a_{n} b_{n}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = \frac{- 1}{12} \\ t_{2} = \frac{- 1}{120} \\ t_{3} = 0 \end{array}$

Explanation of Solution

Given:

$t_{n} = a_{n} b_{n}$ , where $a_{n} = \frac{n - 3}{n!}$ and $b_{n} = a_{n + 3}$

Calculation:

Here, we have

$a_{n} = \frac{n - 3}{n!} \Rightarrow t_{n} = a_{n} b_{n} = (\frac{n - 3}{n!}) (\frac{n + 3 - 3}{(n + 3)!}) = \frac{n - 3}{n!} \frac{n}{(n + 3)!}$

The first three terms of $t_{n} = a_{n} b_{n} = \frac{n - 3}{n!} \frac{n}{(n + 3)!}$ is given by

$\begin{array}{l} t_{1} = \frac{1 - 3}{1!} \frac{1}{(1 + 3)!} = \frac{- 2}{4!} = \frac{- 2}{24} = \frac{- 1}{12} \\ t_{2} = \frac{2 - 3}{2!} \frac{2}{(2 + 3)!} = \frac{- 2}{(2!) (5!)} = \frac{- 2}{240} = \frac{- 1}{120} \\ t_{3} = 0 \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = \frac{- 1}{12} \\ t_{2} = \frac{- 1}{120} \\ t_{3} = 0 \end{array}$

To determine

d)

The first three terms of the sequence having the n^th term is given by $2 a_{n + 1} - 3 a_{n}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = 5 \\ t_{2} = \frac{3}{2} \\ t_{3} = \frac{1}{12} \end{array}$

Explanation of Solution

Given:

$t_{n} = 2 a_{n + 1} - 3 a_{n}$ , where $a_{n} = \frac{n - 3}{n!}$

Calculation:

Here, we have

$\begin{array}{l} a_{n} = \frac{n - 3}{n!} \\ \Rightarrow t_{n} = 2 a_{n + 1} - 3 a_{n} \\ = 2 (\frac{n + 1 - 3}{(n + 1)!}) - 3 (\frac{n - 3}{n!}) \\ = 2 (\frac{n - 2}{(n + 1) n!}) - 3 (\frac{n - 3}{n!}) \\ = \frac{(2 n - 4) - 3 (n - 3) (n + 1)}{(n + 1) n!} \\ = \frac{(2 n - 4) - 3 (n^{2} - 2 n - 3)}{(n + 1)!} \\ ∴ t_{n} = 2 a_{n + 1} - 3 a_{n} = \frac{- 3 n^{2} + 8 n + 5}{(n + 1)!} \end{array}$

The first three terms of $t_{n} = 2 a_{n + 1} - 3 a_{n} = \frac{- 3 n^{2} + 8 n + 5}{(n + 1)!}$ is given by

$\begin{array}{l} t_{1} = \frac{- 3 {(1)}^{2} + 8 (1) + 5}{(1 + 1)!} = \frac{10}{2!} = 5 \\ t_{2} = \frac{- 3 {(2)}^{2} + 8 (2) + 5}{(2 + 1)!} = \frac{9}{3!} = \frac{3}{2} \\ t_{3} = \frac{- 3 {(3)}^{2} + 8 (3) + 5}{(3 + 1)!} = \frac{2}{4!} = \frac{1}{12} \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = 5 \\ t_{2} = \frac{3}{2} \\ t_{3} = \frac{1}{12} \end{array}$

Answer 6

Chapter 11, Problem 1CRE

To determine

a)

The first three terms of the sequence having the n^th term is given by $a_{n}^{2}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of $a_{n}^{2}$ is given by

$\begin{array}{l} a_{1}^{2} = \frac{4}{1!} = 3 \\ a_{2}^{2} = \frac{1}{4} \\ a_{3}^{2} = 0 \end{array}$

Explanation of Solution

Given:

$a_{n} = \frac{n - 3}{n!}$

Calculation:

Here, we have

$a_{n} = \frac{n - 3}{n!} \Rightarrow a_{n}^{2} = {(\frac{n - 3}{n!})}^{2}$

The first three terms of $a_{n}^{2}$ is given by

$\begin{array}{l} a_{1}^{2} = {(\frac{1 - 3}{1!})}^{2} = \frac{4}{1!} \\ a_{2}^{2} = {(\frac{2 - 3}{2!})}^{2} = \frac{1}{(2!) (2!)} = \frac{1}{4} \\ a_{3}^{2} = {(\frac{3 - 3}{3!})}^{2} = 0 \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} a_{1}^{2} = \frac{4}{1!} = 3 \\ a_{2}^{2} = \frac{1}{4} \\ a_{3}^{2} = 0 \end{array}$

Answer 7

Chapter 11, Problem 1CRE

To determine

a)

The first three terms of the sequence having the n^th term is given by $a_{n}^{2}$ .

Answer 8

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of $a_{n}^{2}$ is given by

$\begin{array}{l} a_{1}^{2} = \frac{4}{1!} = 3 \\ a_{2}^{2} = \frac{1}{4} \\ a_{3}^{2} = 0 \end{array}$

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

$a_{n} = \frac{n - 3}{n!}$

Calculation:

Here, we have

$a_{n} = \frac{n - 3}{n!} \Rightarrow a_{n}^{2} = {(\frac{n - 3}{n!})}^{2}$

The first three terms of $a_{n}^{2}$ is given by

$\begin{array}{l} a_{1}^{2} = {(\frac{1 - 3}{1!})}^{2} = \frac{4}{1!} \\ a_{2}^{2} = {(\frac{2 - 3}{2!})}^{2} = \frac{1}{(2!) (2!)} = \frac{1}{4} \\ a_{3}^{2} = {(\frac{3 - 3}{3!})}^{2} = 0 \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} a_{1}^{2} = \frac{4}{1!} = 3 \\ a_{2}^{2} = \frac{1}{4} \\ a_{3}^{2} = 0 \end{array}$

Answer 13

To determine

b)

The first three terms of the sequence having the n^th term is given by $b_{n}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of the sequence is given by

$\begin{array}{l} b_{1} = \frac{1}{4!} \\ b_{2} = \frac{2}{5!} \\ b_{3} = \frac{3}{6!} \end{array}$

Explanation of Solution

Given:

$b_{n} = a_{n + 3}$ , where $a_{n} = \frac{n - 3}{n!}$

Calculation:

Here, we have

$a_{n} = \frac{n - 3}{n!} \Rightarrow b_{n} = a_{n + 3} = (\frac{n + 3 - 3}{(n + 3)!}) = \frac{n}{(n + 3)!}$

The first three terms of $b_{n} = a_{n + 3}$ is given by

$\begin{array}{l} b_{1} = \frac{1}{(1 + 3)!} = \frac{1}{4!} \\ b_{2} = \frac{2}{(2 + 3)!} = \frac{2}{5!} \\ b_{3} = \frac{3}{(3 + 3)!} = \frac{3}{6!} \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} b_{1} = \frac{1}{4!} \\ b_{2} = \frac{2}{5!} \\ b_{3} = \frac{3}{6!} \end{array}$

Answer 14

To determine

b)

The first three terms of the sequence having the n^th term is given by $b_{n}$ .

Answer 15

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of the sequence is given by

$\begin{array}{l} b_{1} = \frac{1}{4!} \\ b_{2} = \frac{2}{5!} \\ b_{3} = \frac{3}{6!} \end{array}$

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

$b_{n} = a_{n + 3}$ , where $a_{n} = \frac{n - 3}{n!}$

Calculation:

Here, we have

$a_{n} = \frac{n - 3}{n!} \Rightarrow b_{n} = a_{n + 3} = (\frac{n + 3 - 3}{(n + 3)!}) = \frac{n}{(n + 3)!}$

The first three terms of $b_{n} = a_{n + 3}$ is given by

$\begin{array}{l} b_{1} = \frac{1}{(1 + 3)!} = \frac{1}{4!} \\ b_{2} = \frac{2}{(2 + 3)!} = \frac{2}{5!} \\ b_{3} = \frac{3}{(3 + 3)!} = \frac{3}{6!} \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} b_{1} = \frac{1}{4!} \\ b_{2} = \frac{2}{5!} \\ b_{3} = \frac{3}{6!} \end{array}$

Answer 20

To determine

c)

The first three terms of the sequence having the n^th term is given by $a_{n} b_{n}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = \frac{- 1}{12} \\ t_{2} = \frac{- 1}{120} \\ t_{3} = 0 \end{array}$

Explanation of Solution

Given:

$t_{n} = a_{n} b_{n}$ , where $a_{n} = \frac{n - 3}{n!}$ and $b_{n} = a_{n + 3}$

Calculation:

Here, we have

$a_{n} = \frac{n - 3}{n!} \Rightarrow t_{n} = a_{n} b_{n} = (\frac{n - 3}{n!}) (\frac{n + 3 - 3}{(n + 3)!}) = \frac{n - 3}{n!} \frac{n}{(n + 3)!}$

The first three terms of $t_{n} = a_{n} b_{n} = \frac{n - 3}{n!} \frac{n}{(n + 3)!}$ is given by

$\begin{array}{l} t_{1} = \frac{1 - 3}{1!} \frac{1}{(1 + 3)!} = \frac{- 2}{4!} = \frac{- 2}{24} = \frac{- 1}{12} \\ t_{2} = \frac{2 - 3}{2!} \frac{2}{(2 + 3)!} = \frac{- 2}{(2!) (5!)} = \frac{- 2}{240} = \frac{- 1}{120} \\ t_{3} = 0 \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = \frac{- 1}{12} \\ t_{2} = \frac{- 1}{120} \\ t_{3} = 0 \end{array}$

Answer 21

To determine

c)

The first three terms of the sequence having the n^th term is given by $a_{n} b_{n}$ .

Answer 22

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = \frac{- 1}{12} \\ t_{2} = \frac{- 1}{120} \\ t_{3} = 0 \end{array}$

Answer 23

Expert Solution

Answer 24

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

$t_{n} = a_{n} b_{n}$ , where $a_{n} = \frac{n - 3}{n!}$ and $b_{n} = a_{n + 3}$

Calculation:

Here, we have

$a_{n} = \frac{n - 3}{n!} \Rightarrow t_{n} = a_{n} b_{n} = (\frac{n - 3}{n!}) (\frac{n + 3 - 3}{(n + 3)!}) = \frac{n - 3}{n!} \frac{n}{(n + 3)!}$

The first three terms of $t_{n} = a_{n} b_{n} = \frac{n - 3}{n!} \frac{n}{(n + 3)!}$ is given by

$\begin{array}{l} t_{1} = \frac{1 - 3}{1!} \frac{1}{(1 + 3)!} = \frac{- 2}{4!} = \frac{- 2}{24} = \frac{- 1}{12} \\ t_{2} = \frac{2 - 3}{2!} \frac{2}{(2 + 3)!} = \frac{- 2}{(2!) (5!)} = \frac{- 2}{240} = \frac{- 1}{120} \\ t_{3} = 0 \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = \frac{- 1}{12} \\ t_{2} = \frac{- 1}{120} \\ t_{3} = 0 \end{array}$

Answer 27

To determine

d)

The first three terms of the sequence having the n^th term is given by $2 a_{n + 1} - 3 a_{n}$ .

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = 5 \\ t_{2} = \frac{3}{2} \\ t_{3} = \frac{1}{12} \end{array}$

Explanation of Solution

Given:

$t_{n} = 2 a_{n + 1} - 3 a_{n}$ , where $a_{n} = \frac{n - 3}{n!}$

Calculation:

Here, we have

$\begin{array}{l} a_{n} = \frac{n - 3}{n!} \\ \Rightarrow t_{n} = 2 a_{n + 1} - 3 a_{n} \\ = 2 (\frac{n + 1 - 3}{(n + 1)!}) - 3 (\frac{n - 3}{n!}) \\ = 2 (\frac{n - 2}{(n + 1) n!}) - 3 (\frac{n - 3}{n!}) \\ = \frac{(2 n - 4) - 3 (n - 3) (n + 1)}{(n + 1) n!} \\ = \frac{(2 n - 4) - 3 (n^{2} - 2 n - 3)}{(n + 1)!} \\ ∴ t_{n} = 2 a_{n + 1} - 3 a_{n} = \frac{- 3 n^{2} + 8 n + 5}{(n + 1)!} \end{array}$

The first three terms of $t_{n} = 2 a_{n + 1} - 3 a_{n} = \frac{- 3 n^{2} + 8 n + 5}{(n + 1)!}$ is given by

$\begin{array}{l} t_{1} = \frac{- 3 {(1)}^{2} + 8 (1) + 5}{(1 + 1)!} = \frac{10}{2!} = 5 \\ t_{2} = \frac{- 3 {(2)}^{2} + 8 (2) + 5}{(2 + 1)!} = \frac{9}{3!} = \frac{3}{2} \\ t_{3} = \frac{- 3 {(3)}^{2} + 8 (3) + 5}{(3 + 1)!} = \frac{2}{4!} = \frac{1}{12} \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = 5 \\ t_{2} = \frac{3}{2} \\ t_{3} = \frac{1}{12} \end{array}$

Answer 28

To determine

d)

The first three terms of the sequence having the n^th term is given by $2 a_{n + 1} - 3 a_{n}$ .

Answer 29

Expert Solution

Answer to Problem 1CRE

Solution:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = 5 \\ t_{2} = \frac{3}{2} \\ t_{3} = \frac{1}{12} \end{array}$

Answer 30

Expert Solution

Answer 31

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given:

$t_{n} = 2 a_{n + 1} - 3 a_{n}$ , where $a_{n} = \frac{n - 3}{n!}$

Calculation:

Here, we have

$\begin{array}{l} a_{n} = \frac{n - 3}{n!} \\ \Rightarrow t_{n} = 2 a_{n + 1} - 3 a_{n} \\ = 2 (\frac{n + 1 - 3}{(n + 1)!}) - 3 (\frac{n - 3}{n!}) \\ = 2 (\frac{n - 2}{(n + 1) n!}) - 3 (\frac{n - 3}{n!}) \\ = \frac{(2 n - 4) - 3 (n - 3) (n + 1)}{(n + 1) n!} \\ = \frac{(2 n - 4) - 3 (n^{2} - 2 n - 3)}{(n + 1)!} \\ ∴ t_{n} = 2 a_{n + 1} - 3 a_{n} = \frac{- 3 n^{2} + 8 n + 5}{(n + 1)!} \end{array}$

The first three terms of $t_{n} = 2 a_{n + 1} - 3 a_{n} = \frac{- 3 n^{2} + 8 n + 5}{(n + 1)!}$ is given by

$\begin{array}{l} t_{1} = \frac{- 3 {(1)}^{2} + 8 (1) + 5}{(1 + 1)!} = \frac{10}{2!} = 5 \\ t_{2} = \frac{- 3 {(2)}^{2} + 8 (2) + 5}{(2 + 1)!} = \frac{9}{3!} = \frac{3}{2} \\ t_{3} = \frac{- 3 {(3)}^{2} + 8 (3) + 5}{(3 + 1)!} = \frac{2}{4!} = \frac{1}{12} \end{array}$

Conclusion:

The first three terms of the sequence is given by

$\begin{array}{l} t_{1} = 5 \\ t_{2} = \frac{3}{2} \\ t_{3} = \frac{1}{12} \end{array}$

Answer 34

Ch. 11.1 - Prob. 1PQ Ch. 11.1 - Prob. 2PQ Ch. 11.1 - Prob. 3PQ Ch. 11.1 - Prob. 4PQ Ch. 11.1 - Prob. 5PQ Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E

a) The first three terms of the sequence having the n th term is given by a n 2 .

Concept explainers

Videos

a)

The first three terms of the sequence having the n^th term is given by $a_{n}^{2}$ .

Answer to Problem 1CRE

Explanation of Solution

b)

The first three terms of the sequence having the n^th term is given by $b_{n}$ .

Answer to Problem 1CRE

Explanation of Solution

c)

The first three terms of the sequence having the n^th term is given by $a_{n} b_{n}$ .

Answer to Problem 1CRE

Explanation of Solution

d)

The first three terms of the sequence having the n^th term is given by $2 a_{n + 1} - 3 a_{n}$ .

Answer to Problem 1CRE

Explanation of Solution

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Chapter 11 Solutions

a) The first three terms of the sequence having the n th term is given by a n 2 .

Concept explainers

Videos

a) The first three terms of the sequence having the nth term is given by an2.

Answer to Problem 1CRE

Explanation of Solution

b) The first three terms of the sequence having the nth term is given by bn.

Answer to Problem 1CRE

Explanation of Solution

c) The first three terms of the sequence having the nth term is given by anbn.

Answer to Problem 1CRE

Explanation of Solution

d) The first three terms of the sequence having the nth term is given by 2an+1−3an.

Answer to Problem 1CRE

Explanation of Solution

Want to see more full solutions like this?

Chapter 11 Solutions

a)

The first three terms of the sequence having the n^th term is given by $a_{n}^{2}$ .

b)

The first three terms of the sequence having the n^th term is given by $b_{n}$ .

c)

The first three terms of the sequence having the n^th term is given by $a_{n} b_{n}$ .

d)

The first three terms of the sequence having the n^th term is given by $2 a_{n + 1} - 3 a_{n}$ .