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 t_{1
}= 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 t_{1} = 3
and r
= 2. Replacing these in the
geometric
formula yields: 3(2)^{n1}
= 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) t_{1}
= 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 

t_{n1}+3(n1) 
From the chart you can see the
formula is t_{n}
= t_{n1} + 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  10^{7})/(1
 10) = 2(9999999)/9 = 2(1111111)
=
2222222
11) Divide numerator
and
denominator by n^{2},
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/(11/4) = 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 n^{2}/(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. 1^{2}
= 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 + . . . + k^{2}
= k(k + 1)(2k + 1)/6 is true.
Prove
it works for n = k + 1
1 + 4 + 9 + . . . + k^{2}
+ (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 + (6k^{2}
+ 12k + 6)/6 =
[(k^{2}
+ k)(2k+1) + (6k^{2}
+ 12k + 6)]/6 =
[(2k^{3}
+ 3k^{2} + k) +
(6k^{2}
+ 12k + 6)]/6 =
(2k^{3}
+ 9k^{2} + 13k +
6)/6
=
Working on the right side gives us:
= [(k^{2}
+ 3k + 2)(2k + 3)]/6
distributing yields:
= (2k^{3}
+ 9k^{2} + 13k +
6)/6
Therfore, it works!!!!
Congratulations!!
You
have finished chapter 13!
Next chapter is
chapter
19!!
Can't wait!!
Bye!