Mathematical Investigations IV
S&S Unit test
Name ______________
You may use a TI-30 Calculator on this exam. Justify all your work.
1.
(2 pts each) Given the numbers 12 and 48, find their:
a) arithmetic mean
b) geometric mean
12  48
 30
2
c) harmonic mean
2 12  48 96

 19.2
12  48
5
12  48  24
2. (4 pts) Consider the series
8
 (1)n1 
k 1
4 12 36 108
  

7 13 19 25

8748
.Write the series in
49
4  3n1
6n  1
79
3. (5 pts) Use appropriate formulas to evaluate the series
 (k  1)
k 5
79
80
k 5
k 6
 (k  1)2   k 2
80
5
k 1
k 1
 k2  k2
80  81161 5  6 11

6
6
 183,825

2
.
  notation.
Mathematical Investigations IV
S&S Unit test
Name ______________
4. (4 pts) The fifth term in an arithmetic sequence is 9 and the 23rd term is 54. Find the sum of
all 23 terms.
d
54  9 5
5
  a1  9  4    1
23  5 2
2
Sum =
5.
n
23
1219
 609.5
 f  l    1  54  
2
2
2
3

(3 pts each) If an   2
 3 an 1
if n  1
if n  1
,
a) State the first five terms of the sequence using common fractions.
4 8 16
3, 2, , ,
3 9 27
b) Give an explicit formula for the sequence.
 2
an  3   
 3
n 1
2k  4
(3 pts) Write out the first four terms of the series 
3k
k 1
92
6.
2
4
 , 0, ,1
3
9
Mathematical Investigations IV
S&S Unit test
Name ______________
n
7. (4 pts) Let S n   ak . Suppose that Sn  n 2  n  2 for all positive integers n > 2 .
k 1
Determine an explicit formula for ak .
an  S n  S n 1
  n 2  n  2    (n  1) 2  (n  1)  2 
  n 2  n  2    n 2  2n  1  n  1  2 
 2n  2
8.
(4 pts) A certain infinite geometric series has only positive terms and its sum is
8
equal to times the arithmetic mean of the first two terms. What is the common ratio
3
of the series.
We are given an infinite geometric series: a  ar  ar 2  .......... 
Also its sum
a
.
1 r
a
8  a  ar 
 
.
1 r 3  2 
Then,
a
8  a  ar 
1
4  1  1r 
 
 


1 r 3  2  1 r 3  1 
 3  4(1  r 2 )
1
4
1
r
2
 r2 
9. (3 pts) Evaluate the following infinite continued fraction: 2 
5
.
5
2
2
5
2
Let x  2 
5
 x  2
5
2
2
5
2
Choose + since x  2
5
5
5
2  4  20
 x 2  2x  5  0  x 
 1 6 .
x
2
Mathematical Investigations IV
S&S Unit test
Name ______________

10.
Let S n represent the nth partial sum of the series
 k 
 ln  k  1  .
k 1
a. (4 pts) Use mathematical induction to prove that Sn   ln(n  1) for all positive integers n.
1
 n 
1
 ln  n  1   ln  2    ln(2) ,
Base Case: (n = 1):
k 1
S1   ln(1  1)   ln(2) , so base case holds.
n
Inductive Step: I Suppose that for some positive integer n,
 k 
 ln  k  1    ln(n  1) .
k 1
Then,
n 1
 k 
 1
2
 n 
 n 1 
ln 
   ln 

 n 1 
 n2
 3
 ln  k  1   ln  2   ln  3   ln  4  
k 1
 n 1 
  ln(n  1)  ln 
 , by induction hypothesis
n2
  ln(n  1)  ln(n  1)  ln( n  2)
  ln(n  2)
Therefore, Sn   ln(n  1) for all positive integers n.

b.(2 pts) Determine whether the series
 k 
 ln  k  1 
converges or diverges. If it converges,
k 1
find its value. You may use the formula for the partial sum given in part a..

 k 
 ln  k  1   lim  S   lim ln(n  1) , which diverges! So series diverges.
k 1
n 
n
n 