Mathematical Investigations IV
Name
Mathematical Investigations IV
Iteration Forever
SERIES
A series is the sum of the terms in a sequence.
For example:
Sequence
Corresponding Series
2, 7, 12, 17, 22, 27, 32
2 + 7 + 12 + 17 + 22 + 27 + 32
, , , 
+++
1, 1, 2, 3, 5, 8
1+1+2+3+5+8
To aid in describing series, we note this with the capital Greek letter sigma, .
For a sequence ak  n
k 1
n
, the corresponding series may be written in the form
 ak .
k 1
For example, the explicit formula for the first sequence ak k 1 above is ak  5k  3 .
7
7
The corresponding series is defined by
  5k  3 .
k 1
1 – 4. Expand each series and find its sum.
7
  5k  3
1.
5
=
2.
k 1
 n 1 

 n  =

n 1  2
n2
6
3.

20
4.
(1) n
=
3n
1
1 
  n  n  1  =
n 1
Seq & Ser. 7.1
Rev. F07
Mathematical Investigations IV
Name
5 – 10. Write each series in  - notation.
5.
5 + 10 + 15 + … + 100
6.
–9 – 2 + 5 + … + 369
7.
1·3 + 2·5 + 3·7 + … + 49·99
8.
1 – 3 + 5 – 7 + … + 49
9.
1 12  13  14  15 
10.
1
 101
1
2
16
25
49
64
 44  89  16
 32
 36
64  128  256
Sequence of Partial Sums
11. Consider the sequence defined by ak  4k  2 .
a. List ak  4
k 1
b. We want to create a new sequence S1 , S2 , S3 , S4 associated with the sequence ak k 1 .
4
n
The sequence S n n 1 is called the sequence of partial sums, where S n   ak .
4
k 1
In other words, we have the following
for the specific sequence above:
for the general case:
S1  6  6
S1  a1
S2  6  10  16
S 2  a1  a2
S3  6  10  14  30
S3  a1  a2  a3
S4  6  10  14  18  48
S 4  a1  a2  a3  .a4
Again, more generally, the sequence of partial sums is given by
Sn 
n
where S n   ak  a1  a2  a3  ...  an
k 1
c. List S n n 1
4
12.
Let bk  5  3k .
Seq & Ser. 7.2
Rev. F07
Mathematical Investigations IV
Name
a. Find bk k 1 .
5
n
b. Find S k k 1 where Sn   bk is the associated sequence of partial sums.
5
k 1
13. Given a general sequence ak k 1 , find a formula to express S n n 1 recursively.
8
8
Special Sums
14. Though we often use the notation ak, we sometimes refer to a(k) just as we use the
notation ƒ(x). This reminds us that the sequence is a function of k.
Let ak be the constant sequence given by ak  a(k )  3 . Find each of the following terms.
a(1) =
, a(2) =
, a(5) =
, a(50) =
, a(500) =
5
Consider
 3.
Expand the series (show the terms) and then find the sum.
k 1
15. Expand each series without simplifying. (Use "…" in part b.)
6
a.
n
4 
b.
j1
4 
j1
n
16. A special sum: Write out the sum
c
in terms of c, where c is a constant (do not
j 1
simplify). Then find the sum.
n
Sn   c 
j 1
Seq & Ser. 7.3
Rev. F07
Mathematical Investigations IV
Name
17. Evaluate
12
17
 18
a.
b.
k 1
 (3)
k 4
n
Another Special Series: Sn   j
18.
j 1
n
Let Sn   j . Complete the first three and the last three terms of this sum.
j 1
Sn =
+
+
+... +
+
+
Complete this sum again, but reverse the order of the terms. Fill in each blank.
Sn =
+
+
+... +
+
+
Add vertically the two previous equations. That is, add the two terms in the first
blanks, then the two in the second blanks, etc., and simplify.
2Sn =
+
+
+... +
+
+
Simplify:
2·Sn =
Solve for Sn, giving the handy formula,
n
Sn   j 
j 1
19. Find:
50
a.
 j.
100
b.
j 1
 (k  3) =
k 1
100
c.
40
e.
 (3k ) =
k 1
Seq & Ser. 7.4
j
j  51
j 1
20
d.
j
n
f.
 (2k  1) =
k 1
Rev. F07
Mathematical Investigations IV
Name
20. Show:
n
a
n
n
 c  a j  c a j
a.
j 1
b.
j 1
j 1
j  b j   a j   b j
n
n
j 1
j 1
21. One more special series: Arithmetic Series
We just considered one specific arithmetic sequence, but we want to generalize that
formula. First, write out the formula for the nth term of an arithmetic sequence:
an =
Fill in the blanks, using the same approach as you used on the previous page.
n
Let Sn =
  a  (k  1)  d  .
In expanded notation, we can write the terms:
k 1
Sn =
+
+
+...+
+
+
.
+...+
+
+
.
Write that sum again, but in reverse order:
Sn =
+
+
Now, add the two previous equations:
2Sn=
+
+
Simplify: 2Sn =
+...+
+
+
.
.
This results in the general formula for an arithmetic series: Sn =
.
Now if we let a be the first term and be the last or nth term, then
a second, but very nice, formula for Sn:
n
Sn   a  
2
22. Evaluate the following.
= a + (n –1)d and we get
10
a.
  3k  2  =
b. Find n if a1 = 5, d = 3, and an = 68.
k 1
c.
–2 – 1.7 – 1.4 – ... + 11.5 =
d.
Seq & Ser. 7.5
Find
if S20 = 153 and a = 0.7.
Rev. F07
Mathematical Investigations IV
Name
23. Yet one more: Geometric Series
Let's turn our attention to geometric series. First, write the formula for the nth term of a
geometric sequence.
an =
Now for the formula for a geometric series:
  ar  .
n
Let Sn =
k 1
In expanded notation,
k 1
Sn =
+
+
+...+
+
.
Multiply each side of the previous equation by r and vertically line up similar terms.
r Sn =
+
+ ...+
+
+
.
Subtract the 2nd equation from the 1st equation:
Sn – r Sn =
.
Factor out Sn from the left side and factor out a from the right side.
Solving the previous equation for Sn yields:
Sn =
In short, the sum of the first n terms in a geometric series is Sn 
a 1  r n 
1 r
.
24. Try a few exercises using this formula: (Begin by identifying the first term and the common ratio.)
  3  k 1 
a.  12     =
 2 
k 1 

b.
c. Find n if Sn = 51010.0501,
d. Find a if S7 
10
1 – 2 + 4 – 8 + 16 – . . .  32768 =
18394
4
and r 
729
3
a1 = 10000 and a2 = 10100
Seq & Ser. 7.6
Rev. F07