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Chapter 12 Sequences; Induction; the Binomial Theorem

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<strong>Chapter</strong> <strong>12</strong>: <strong>Sequences</strong>; <strong>Induction</strong>; <strong>the</strong> <strong>Binomial</strong> <strong>Theorem</strong><br />

c. Scroll through <strong>the</strong> table:<br />

(d) Scroll through <strong>the</strong> table:<br />

At <strong>the</strong> beginning of <strong>the</strong> 33rd month, or after<br />

32 payments have been made, <strong>the</strong> balance is<br />

below $4,000. The balance is $3982.8.<br />

d. Scroll through <strong>the</strong> table:<br />

At <strong>the</strong> beginning of <strong>the</strong> 192nd month, or<br />

after 191 payments have been made, <strong>the</strong><br />

balance is below $100,000. The<br />

balance is about $99,288.<br />

(e) Scroll through <strong>the</strong> table:<br />

The loan will be paid off after 289<br />

months. They will make 288 payments<br />

of $1332.34 plus a last payment of<br />

$1142.80 plus interest. The total<br />

amount paid is:<br />

⎛ 0.0675 ⎞<br />

288(1332.34) + 1142.80⎜1+<br />

⎟<br />

⎝ <strong>12</strong> ⎠<br />

≈ $384,863.15<br />

(f) The total interest expense is <strong>the</strong><br />

difference of <strong>the</strong> total of <strong>the</strong> payments<br />

and <strong>the</strong> original loan:<br />

384,863.15 − 190,000 = $19,863.15 .<br />

69. a. b1 = 5,000, bn<br />

= 1.015bn<br />

− 1−<br />

100<br />

b2 = 1.015b( 2−1)<br />

− 100 = 1.015b1−100<br />

= 1.015 5000 − 100 = $4975<br />

( )<br />

b. Enter <strong>the</strong> recursive formula in Y= and draw<br />

<strong>the</strong> graph:<br />

The balance will be paid off at <strong>the</strong> end of 94<br />

months or 7years and 10 months.<br />

Total payments =<br />

100(93) + 11.01 1.015 = $9,311.18<br />

( )<br />

e. The total interest expense is <strong>the</strong> difference<br />

between <strong>the</strong> total of <strong>the</strong> payments and <strong>the</strong><br />

original balance:<br />

100(93) + 11.18 − 5,000 = $4,311.18<br />

70. This is a geometric sequence with<br />

a1 = 20,000, r = 1.04, n = 5 . Find <strong>the</strong> fifth term<br />

of <strong>the</strong> sequence:<br />

5−1 4<br />

a 5 = 20,000(1.04) = 20,000(1.04) = 23,397.17<br />

Her salary in <strong>the</strong> fifth year will be $23,397.17.<br />

<strong>Chapter</strong> <strong>12</strong> Test<br />

1.<br />

2<br />

n −1<br />

an<br />

=<br />

n + 8<br />

2 2<br />

1 −1 0 2 −1 3<br />

a1 = = = 0, a2<br />

= = ,<br />

1+ 8 9 2+<br />

8 10<br />

2 2<br />

3 −1 8 4 −1 15 5<br />

a3 = = , a4<br />

= = = ,<br />

3+ 8 11 4+<br />

8 <strong>12</strong> 4<br />

2<br />

5 −1 24<br />

a5<br />

= =<br />

5+<br />

8 13<br />

The first five terms of <strong>the</strong> sequence are 0, 3<br />

10 ,<br />

8<br />

11 , 5 24<br />

, and<br />

4 13 .<br />

<strong>12</strong>80<br />

© 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as <strong>the</strong>y currently<br />

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from <strong>the</strong> publisher.

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