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 10, 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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A circle of radius r centered at the point (0,r) in the plane will intersect the y-axis at the origin and the point A=(0,2r), as pictured below. A line passes through the point A and the point C=(11/2,0) on the x-axis. In this problem, we will investigate the coordinates of the intersection point B between the circle and the line, as 1 → ∞ A=(0,2r) B (0,0) (a) The line through A and C has equation: y= 2 117 x+27 (b) The x-coordinate of the point B is 4472 121,2 +4 40 (c) The y-coordinate of the point B is +27 121 44 (d) The limit as r→ ∞ of the x-coordinate of B is 121 (if your answer is oo, write infinity).

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Answer 2

Question

Chapter 10, 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 10, 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 10, 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 10, 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 10, 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 10, 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. 10.1 - Prob. 1PQ Ch. 10.1 - Prob. 2PQ Ch. 10.1 - Prob. 3PQ Ch. 10.1 - Prob. 4PQ Ch. 10.1 - Prob. 5PQ Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E

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

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

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Chapter 10 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 $2 a_{n + 1} - 3 a_{n}$ .