Chapter 12 Sequences; Induction; the Binomial Theorem

Section 12.1: Sequences
16 16 16
2 2
∑ ∑ ∑
77. ( k )
+ 4 = k + 4
k= 1 k= 1 k=
1
( + )( ⋅ + )
16 16 1 2 16 1
= + 416
6
= 1496 + 64 = 1560
( )
14 14
78. ∑ ( k
2 − 4) = ( 0 2 − 4) + ∑( k
2 −4)
79.
k= 0 k=
1
14 14
2
∑
= k −
∑
k= 1 k=
1
4
( + )( ⋅ + )
14 14 1 2 14 1
=− 4+ −4 14
6
=− 4 + 1015− 64 = 955
⎡ ⎤
2k = 2 2k = 2⎢
k − k⎥
⎣ ⎦
60 60 60 9
∑ ∑ ∑ ∑
k= 10 k= 10 k= 1 k=
1
( + ) ( + )
⎡60 60 1 9 9 1 ⎤
= 2 ⎢ − ⎥
⎣ 2 2 ⎦
= 2 1830 − 45 = 3570
[ ]
⎡ ⎤
k k ⎢ k k⎥
⎣ ⎦
40 40 40 7
80. ∑( − 3 ) =− 3∑ =−3
∑ −∑
k= 8 k= 8 k= 1 k=
1
( + ) ( + )
⎡40 40 1 7 7 1 ⎤
=−3⎢
− ⎥
⎣ 2 2 ⎦
=−3 820 − 28 =−2376
[ ]
( )
From the table we see that the balance is
below $2000 after 14 payments have been
made. The balance then is $1953.70.
c. Scrolling down the table, we find that
balance is paid off in the 36th month. The
last payment is $83.78. There are 35
payments of $100 and the last payment of
$83.78. The total amount paid is: 35(100) +
83.78(1.01) = $3584.62. (we have to add
the interest for the last month).
d. The interest expense is:
3584.62 – 3000.00 = $584.62
81.
82.
20 20 4
3 3 3
∑ ∑ ∑
k = k − k
k= 5 k= 1 k=
1
( + ) ( + )
2 2
2 2
⎡20 20 1 ⎤ ⎡4 4 1 ⎤
= ⎢ ⎥ −⎢ ⎥
⎣ 2 ⎦ ⎣ 2 ⎦
= 210 − 10 = 44,000
24 24 3
3 3 3
∑ ∑ ∑
k = k − k
k= 4 k= 1 k=
1
( + ) ( + )
2 2
2 2
⎡24 24 1 ⎤ ⎡3 3 1 ⎤
= ⎢ ⎥ −⎢ ⎥
⎣ 2 ⎦ ⎣ 2 ⎦
= 300 − 6 = 89,964
84. a. B 1 = 1.005(18500) − 534.47 = $18,058.03
b. Put the graphing utility in SEQuence mode.
Enter Y= as follows, then examine the
TABLE:
83. a. B 1 = 1.01(3000) − 100 = $2930
b. Put the graphing utility in SEQuence mode.
Enter Y= as follows, then examine the
TABLE:
From the table we see that the balance is
below $10,000 after 19 payments have been
made. The balance then is $9713.76.
1239
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