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Chapter 19

1.
1. $a_{1} = 8, d = - 1/2;$ $a_{1} = 8, d = - 1/2;$ arith. seq. 8, 15/2, 7, 13/2, 6, …
2. $a_{1} = 8, r = - 1/2;$ $a_{1} = 8, r = - 1/2;$ geom. seq. 8, $- 4, 2, - 1, 1/2, \dots$ $- 4, 2, - 1, 1/2, \dots$
2. 6, $- 2, \frac{2}{3}, \dots;$ $- 2, \frac{2}{3}, \dots;$

geometric sequence

$\begin{array}{rcl} a_{1} & = & 6 r = \frac{- 2}{6} = - \frac{1}{3} \\ S_{7} & = & \frac{6 [1 - {(- \frac{1}{3})}^{7}]}{1 - (- \frac{1}{3})} \\ = & \frac{6 (1 + \frac{1}{3^{7}})}{\frac{4}{3}} = \frac{1094}{243} \end{array}$ $\begin{array}{rcl} a_{1} & = & 6 r = \frac{- 2}{6} = - \frac{1}{3} \\ S_{7} & = & \frac{6 [1 - {(- \frac{1}{3})}^{7}]}{1 - (- \frac{1}{3})} \\ = & \frac{6 (1 + \frac{1}{3^{7}})}{\frac{4}{3}} = \frac{1094}{243} \end{array}$
3. $\begin{array}{l} d = (x + 6) - 4 & and & d = (3 x + 2) - (x + 6) \\ d = x + 2 & and & d = 2 x - 4 \end{array}$ $\begin{array}{l} d = (x + 6) - 4 & and & d = (3 x + 2) - (x + 6) \\ d = x + 2 & and & d = 2 x - 4 \end{array}$

$\begin{array}{rcl} So x + 2 & = & 2 x - 4 \\ x & = & 6 \end{array}$ $\begin{array}{rcl} So x + 2 & = & 2 x - 4 \\ x & = & 6 \end{array}$

$\begin{array}{rcl} Thus, d & = & 8. \\ a_{10} & = & 4 + 9 (8) = 76 \\ S_{10} & = & \frac{10}{2} (4 + 76) = 400 \end{array}$ $\begin{array}{rcl} Thus, d & = & 8. \\ a_{10} & = & 4 + 9 (8) = 76 \\ S_{10} & = & \frac{10}{2} (4 + 76) = 400 \end{array}$
4. $\begin{array}{l} 0.454545 \dots = 0.45 + 0.0045 + 0.000045 + \dots \\ a = 0.45 r = 0.01 \\ S = \frac{0.45}{1 - 0.01} = \frac{0.45}{0.99} = \frac{5}{11} \end{array}$ $\begin{array}{l} 0.454545 \dots = 0.45 + 0.0045 + 0.000045 + \dots \\ a = 0.45 r = 0.01 \\ S = \frac{0.45}{1 - 0.01} = \frac{0.45}{0.99} = \frac{5}{11} \end{array}$
5. $\begin{array}{rcl} \sqrt{1 - 4 x} & = & {(1 - 4 x)}^{1/2} \\ = & 1 + (\frac{1}{2}) (- 4 x) + \frac{\frac{1}{2} (\frac{1}{2} - 1)}{2} {(- 4 x)}^{2} + \dots \\ = & 1 - 2 x - 2 x^{2} + \dots \end{array}$ $\begin{array}{rcl} \sqrt{1 - 4 x} & = & {(1 - 4 x)}^{1/2} \\ = & 1 + (\frac{1}{2}) (- 4 x) + \frac{\frac{1}{2} (\frac{1}{2} - 1)}{2} {(- 4 x)}^{2} + \dots \\ = & 1 - 2 x - 2 x^{2} + \dots \end{array}$
6. $\begin{array}{l} {(2 x - y)}^{5} = {(2 x)}^{5} + 5 {(2 x)}^{4} (- y) + \frac{5 (4)}{2} {(2 x)}^{3} {(- y)}^{2} \\ + \frac{5 (4) (3)}{2 (3)} {(2 x)}^{2} {(- y)}^{3} + \frac{5 (4) (3) (2)}{2 (3) (4)} (2 x) {(- y)}^{4} + {(- y)}^{5} \\ = 32 x^{5} - 80 x^{4} y + 80 x^{3} y^{2} - 40 x^{2} y^{3} + 10 x y^{4} - y^{5} \end{array}$ $\begin{array}{l} {(2 x - y)}^{5} = {(2 x)}^{5} + 5 {(2 x)}^{4} (- y) + \frac{5 (4)}{2} {(2 x)}^{3} {(- y)}^{2} \\ + \frac{5 (4) (3)}{2 (3)} {(2 x)}^{2} {(- y)}^{3} + \frac{5 (4) (3) (2)}{2 (3) (4)} (2 x) {(- y)}^{4} + {(- y)}^{5} \\ = 32 x^{5} - 80 x^{4} y + 80 x^{3} y^{2} - 40 x^{2} y^{3} + 10 x y^{4} - y^{5} \end{array}$
7. $5 % = 0.05$ $5 % = 0.05$

Value after 1 year is

$2500 + 2500 (0.05) = 2500 (1.05)$ $2500 + 2500 (0.05) = 2500 (1.05)$

$\begin{array}{rcl} V_{20} & = & 2500 {(1.05)}^{20} \\ = & $ 6633.24 \end{array}$ $\begin{array}{rcl} V_{20} & = & 2500 {(1.05)}^{20} \\ = & $ 6633.24 \end{array}$
8. $2 + 4 + \dots + 200$ $2 + 4 + \dots + 200$

$a_{1} = 2, a_{100} = 200, n = 100$ $a_{1} = 2, a_{100} = 200, n = 100$

$S_{100} = \frac{100}{2} (2 + 200) = 10, 100$ $S_{100} = \frac{100}{2} (2 + 200) = 10, 100$
9. Ball falls 8.00 ft, rises 4.00 ft, falls 4.00 ft, etc.

$\begin{array}{rcl} Distance & = & 8.00 + (4.00 + 4.00) + (2.00 + 2.00) + \dots \\ = & 8.00 + 8.00 + 4.00 + 2.00 + \dots \\ = & 8.00 + \frac{8.00}{1 - 0.5} = 8.00 + 16.0 = 24.0 ft \end{array}$ $\begin{array}{rcl} Distance & = & 8.00 + (4.00 + 4.00) + (2.00 + 2.00) + \dots \\ = & 8.00 + 8.00 + 4.00 + 2.00 + \dots \\ = & 8.00 + \frac{8.00}{1 - 0.5} = 8.00 + 16.0 = 24.0 ft \end{array}$

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Table of Contents for Chapter 19

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Table of Contents for
Chapter 19