Here are the answers
worked out for the sample test!
1) Simply replace n with 1, 2, 3, and 4.
The terms are -8, 16, -32, 64.
The sequence is geometric with r = -2
2) Do the same as #1 and get the sequence: 4,
9, 14, 19. This
sequence is arithmetic
with d = 5.
3) Do the same as #1 and get the sequence:
4, 7, 12, 19 which is
neither arithmetic or geometric.
4) This sequence is arithmetic with t1
= 5 and d = -4. Replacing these
values into the arithmetic formula gives us 5 + (n - 1)(-4) = 5 -4n +4
= 9 - 4n.
5) The sequence is geometric
with t1 = -3 and r
= 2. Replacing these in the geometric
formula yields: -3(2)n-1
= -3(2)n(2)-1
= -3/2(2)n.
6) This recursive definition says to add the
previous two terms to get the next term.
Therefore, the terms would be: 5, 8, 13, 21, 34.
7) This sequence says to
double the previous term and add the current term number. Thus
the sequence is: 3, 8, 19, 42, 89.
8) t1
= 2 Look at the following table to figure out the definition:
| term |
think |
pattern |
| 1 |
2 |
2 |
| 2 |
2+3 |
2+3(1) |
| 3 |
5+6 |
5+3(2) |
| 4 |
11+9 |
11+3(3) |
| 5 |
20+12 |
20+3(4) |
| 6 |
32+15 |
32+3(5) |
| n |
|
tn-1+3(n-1) |
From the chart you can see the
formula is tn
= tn-1 + 3(n - 1)
9) To find the sum you need the first term and
the last term. The first term is 5 and the 35th term
is 107. Put these in the formula and get
S = 35(5 + 107)/2 = 35(112)/2 = (35)(56) = 1960
10) This is geometric with
r = 10. Substitute the values in the formula and get:
S = 2(1 - 107)/(1
- 10) = 2(-9999999)/-9 = 2(1111111) =
2222222
11) Divide numerator and
denominator by n2,
which is the highest power
in the denominator. = (2 + 0 - 0)/(1 - 0) = 2/1 = 2
12) The lim of 1/n is 0.
Thus tan 0 = 0
13) Look at what kind of
number you are raising to the n power. It's less
than 1. thus the limit = 0
14) This is a geometric series with r = 1/4.
Therefore, it converges to the following sum: =
1/(1-1/4) = 1/3/4 = 4/3
15) This is also a geometric
series with r = 2. Therefore, it diverges.
16) r = x. From the definition, this will
converge if |x| < 1. Using our knowledge of
absolute value, gives us: -1 < x < 1, the interval
of convergence.
17) Think of it as .37 + .0037 + .000037 + .
. . r = .01 Using the
formula gives us: .37/(1 - .01) = .37/.99 = 37/99
18) The series follows the pattern of n2/(n+1)2
It has 6 terms. So the sigma notation is: 
19) Simply replace k one at a time with 1, 2,
3, 4, 5, 6, 7, 8. The result is: 3
+ 8 + 15 + 24 + 35 + 48 + 63 + 80
20) Prove it works for n = 1. 12
= 1(1 + 1)(2(1) +1)/6
1 = 2(3)/6 = 1 It works for n = 1
Assume it
works for n = k. That is assume the following:
1 + 4 + 9 + 16 + . . . + k2
= k(k + 1)(2k + 1)/6 is true.
Prove
it works for n = k + 1
1 + 4 + 9 + . . . + k2
+ (k + 1)2 = (k+1)(k+1+1)(2k+2+1)/6
replace the sequence in red with what you assumed
to be true in step 2.
k(k+1)(2k+1)/6 +(k+1)2
= (k+1)(k+2)(2k+3)/6
Working on the left side yields:
k(k+1)(2k+1)/6 + (6k2
+ 12k + 6)/6 =
[(k2
+ k)(2k+1) + (6k2
+ 12k + 6)]/6 =
[(2k3
+ 3k2 + k) + (6k2
+ 12k + 6)]/6 =
(2k3
+ 9k2 + 13k + 6)/6
=
Working on the right side gives us:
= [(k2
+ 3k + 2)(2k + 3)]/6
distributing yields:
= (2k3
+ 9k2 + 13k + 6)/6
Therfore, it works!!!!
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Congratulations!! You
have finished chapter 13!
Next chapter is chapter
19!!
Can't wait!!
Bye!
