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MTH001
Elementary Mathematics
LECTURE #
14
SERIES:
Thesum
of the terms of a sequence
forms a series. If a1,
a2, a3, ...
represent
a sequence of numbers,
thenthe corresponding series
is
∞
∑a
=
a1 +
a2 + a3
+ ...
k
k
=1
SUMMATIONNOTATION
Thecapital
Greek letter sigma∑
is
used to write a sum in a
short hand notation.
where
k varies from 1 to n
representsthe sum given in
expandedform by
=
a1 +
a2 + a3
+ ... + an
Moregenerally
if m and n areintegers and m
≤
n,
then the summation from k
equal m to n of
ak is
n
∑a
=
am +
am+1
+
am+2
+
L
+
an
k
k
=m
Herek is
calledthe index of the
summation;m
thelower
limit of the summationand
n
the
upperlimit
of thesummation.
COMPUTINGSUMMATIONS:
Let
a0 =
2, a1 = 3, a2
= -2,
a3 = 1 and a4
=
0.Computeeach of
thesummations:
4
2
1
∑a
∑a
∑a
1.
2.
2j
i
k
i=0
j=0
k
=1
SOLUTION:
4
∑
=2
(
1.
ai = a0++3a+
+-2)2
+ 13++0a4=
4
a
+a
1
i
=0
2
2
∑a
2j =a
+a +a
0
2
4
j
=0
=
2 + (-2) + 0 = 0
1
∑a
k
3.
=
a1
k
=1
=3
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Elementary Mathematics
EXERCISE:
Computethe
summations
3
∑
(2i -
1)
=
[2(1)- 1] +
[2(2)- 1] +
[2(3)- 1]
1.
i=1
=
1
+
3+
5
=
9
1
∑
(k
+
2)
=
[(-1)3 +
2]
+
[(0)3 + 2] +
[(1)3 + 2]
3
2.
k
=-1
=
[-1
+
2]
+
[0
+
2]
+
[1
+
2]
=
1+ 2+
3
=
6
SUMMATIONNOTATION
TO EXPANDEDFORM:
(-1)i
n
Writethe
summation ∑
to
expanded form:
i=0 i
+
1
SOLUTION:
(-1)i
(-1)0 (-1)1 (-1)2 (-1)3
(-1)n
n
∑
i
+1 =
0
+1
+
1+1 +
2
+1
+
3
+1
+L
+
n
+1
i=0
1
(-1)
1 (-1)
(-1)n
=+
++
+L +
n
+1
1
2
3
4
(-1)n
111
=
1-
+ - +L +
n
+1
234
EXPANDEDFORM
TO SUMMATIONNOTATION:
Writethe
following usingsummation
notation:
n
+1
1
2
3
+
+
+L +
1.
n
n +1
n
+
2
2n
SOLUTION:
We
find the kth term of
the series.
Thenumerators
forms an arithmeticsequence 1, 2, 3,...,
n+1, in which
a
= first term = 1
&
d
= common difference = 1
ak = a + (k -
1)d
=
1 + (k - 1) (1) = 1 + k - 1 = k
Similarly,the
denominators forms an arithmetic
sequence
n,
n+1, n+2, ..., 2n, in
which
a
= first term = n
d
= common difference = 1
∴
ak = a + (k -
1) d
=
n + (k - 1) (1)
=k+n-1
Hencethe
kth term of theseries
is
k
(n -
1)
+
k
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MTH001
Elementary Mathematics
And
the expression for
theseries is given by
n
+
1
n+1
1
2
3
k
=∑
∴
+
+
+L +
n
n +1
n
+
2
k
=1 (n
-
1)
+
k
2n
k
+1
n
=∑
k
=0 n
+
k
TRANSFORMING
A SUM BY A CHANGE OF
VARIABLE:
3
∑k
Consider
=
12 +
22 + 32
2
k
=1
3
∑i
=
12 +
22 + 32
2
and
i=1
3
3
∑k
=
∑ i2
2
Hence
k
=1
i=1
Theindex
of a summation can be replaced by
any othersymbol. The
index of a summation
is
therefore called a
dummyvariable.
EXERCISE:
n+1
k
Consider
∑
(n -
1)
+
k
k
=1
Substituting
k = j + 1 so that j = k 1
When
k = 1, j = k - 1 = 1 - 1 = 0
When
k = n + 1, j = k - 1 = (n + 1) - 1 = n
Hence
j
+1
n+1
n
k
∑
(n -
1)
+
k
=
∑
(n -
1)
+
(
j
+
1)
k
=1
j=0
j
+1 n
k +1
n
=∑
=∑
(changingvariable)
j=0 n
+
j
k
=0 n
+
k
Transform
by making the change of
variable j = i - 1, in
thesummation
n-1
i
∑
(n -
i)
**
2
i=1
PROPERTIES
OF SUMMATIONS:
n
n
n
∑
(a
+
bk ) =
∑ ak + ∑ bk ;
ak , bk ∈
R
1.
k
k
=m
k
=m
k
=m
n
n
∑
ca
=
c
∑ ak
2.
c∈R
k
k
=m
k
=m
b-i
b
∑
(k +
i)
=
∑ k
i∈N
3.
k
=a-i
k
=a
b+i
b
∑
(k -
i)
=
∑ k
i∈N
4.
k
=a+i
k
=a
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MTH001
Elementary Mathematics
n
∑c
=
c
+
c
+
L
+
c
=
nc
5.
k
=1
EXERCISE:
Expressthe
following summation
moresimply:
n
n
3∑
(2k -
3)
+
∑ (4
-
5k )
k
=1
k
=1
SOLUTION:
n
n
3∑
(2k -
3)
+
∑ (4
-
5k )
k
=1
k
=1
n
n
=
3∑
3(2k -
3)
+
∑ (4
-
5k )
k
=1
k
=1
n
=
∑ [3(2k -
3)
+
(4
-
5k )]
k
=1
n
=
∑ (k -
5)
k
=1
n
n
=
∑k
-
∑5
k
=1
k
=1
n
=
∑ k
-
5n
k
=1
ARITHMETICSERIES:
Thesum
of the terms of an arithmetic
sequence forms an arithmetic
series (A.S).For
example
1+3+5+7+...
is
an arithmetic series of
positiveodd integers.
In
general, if a is the
firstterm and d the
commondifference of an arithmetic
series,then the
series
is given as: a + (a+d) +
(a+2d) +...
SUM
OF n TERMS OF AN
ARITHMETICSERIES:
Let
a be the first term and d
be the common difference of an
arithmetic series. Then
itsnth
termis:
an = a + (n -
1)d; n ≥
1
If
Sn denotes the sum of
first n terms of the
A.S,then
Sn = a + (a +
d) + (a + 2d) + ... + [a + (n-1)
d]
=
a + (a+d) + (a + 2d) + ... + an
=
a + (a+d) + (a + 2d) + ... + (an - d) +
an .........(1)
where
an = a + (n - 1) d
Rewritingthe
terms in the series in
reverse order,
Sn
=
an + (an
- d) + (an - 2d) + ... + (a + d) + a
..........(2)
Adding(1)
and (2) term by term,
gives
2
Sn = (a + an) + (a +
an) + (a + an) + ... + (a
+ an)
(n
terms)
2
Sn = n (a + an)
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MTH001
Elementary Mathematics
⇒
Sn =
n(a + an)/2
Sn =
n(a + l)/2.......................(3)
Where
l
= an = a + (n - 1)d
Therefore
Sn=
n/2 [a + a + (n - 1) d]
Sn n/2 [2 a + (n - 1)
d]..........(4)
EXERCISE:
Findthe
sum of first n
naturalnumbers.
SOLUTION:
Let
Sn =
1 + 2 + 3 + ... + n
Clearlythe
right hand sideforms an
arithmetic serieswith
a
= 1, d = 2 - 1 = 1
and
n = n
n
[
2a +
(n -
1)d ]
∴
=
Sn
2
n
[
2(1)+ (n -
1)(1)]
=
2
n
[
2
+
n
-
1]
=
2
n(n +
1)
=
2
EXERCISE:
Findthe
sum of all twodigit
positive integerswhich are
neither divisible by 5 nor by
2.
SOLUTION:
Theseries
to be summedis:
11
+ 13 + 17 + 19 + 21 + 23 + 27 + 29 + ... + 91 + 93 + 97 +
99
which
is not an
arithmeticseries.
If
we make group of four terms
we get
(11
+ 13 + 17 + 19) + (21 + 23 + 27 + 29) +
(31 + 33 + 37 + 39) + ... + (91 + 93 + 97
+ 99) =
60
+ 100 + 140 + ... +
380
whichnow
forms an arithmeticseries in
which
a
= 60; d = 100 - 60 = 40
and
l
= an = 380
To
find n, we use
theformula
an
=
a
+ (n - 1) d
⇒
380
=
60
+ (n - 1) (40)
⇒
380 - 60
=
(n
- 1) (40)
⇒
320
=
(n
- 1) (40)
320
=
n
-1
40
8
=n-1
⇒
n
=9
Now
n
=
(a +
l
)
Sn
2
9
∴
=
(60+ 380)
=
1980
S9
2
GEOMETRICSERIES:
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MTH001
Elementary Mathematics
Thesum
of the terms of a geometric
sequence forms a geometric
series (G.S.).For
example
1
+ 2 + 4 + 8 + 16 + ...
is
geometric series.
In
general, if a
is
thefirst term and
r
thecommon
ratio of a geometricseries,
then theseries
is
given as: a + ar + ar2 +
ar3 + ...
SUM
OF n TERMS OF A
GEOMETRICSERIES:
Leta be
thefirst term and
r
be
thecommon ratio of a
geometricseries. Then its
nthterm is:
an =
arn-1; n ≥
1
If
Sn denotes the sum of first
n terms of the
G.S.then
Sn = a + ar +
ar2 + ar3
+ ... + arn-2 + arn-1...............(1)
Multiplyingboth
sides by r we get.
r
Sn =
ar + ar2 + ar3
+ ... + arn-1 + arn..................(2)
Subtracting(2)
from (1) we get
Sn -
rSn= a arn
(1
- r) Sn = a (1 -
rn)
⇒
a(1
-
r
n )
⇒
Sn =
(r ≠
1)
1-
r
EXERCISE:
Findthe
sum of the
geometricseries
22
6-2+
-
+ L
+
to 10
terms
39
SOLUTION:
In
the given
geometricseries
-2
1
a
=
6,
r
=
=-
and
n
=
10
6
3
a(1
-
r
n )
∴
=
Sn
1-
r
⎛
⎛ 1
⎞10
⎞
6
⎜1
- ⎜ -
⎟ ⎟ 6
⎛1
+
1
⎞
⎜
⎝ 3⎠
⎟
⎜
10
⎟
⎠=
⎝ 3
⎠
⎝
=
S10
⎛
1⎞
⎛
4⎞
1- ⎜ -
⎟
⎜⎟
⎝
3⎠
⎝
3⎠
⎛
1⎞
9
⎜1
+
10
⎟
= ⎝
3 ⎠
2
INFINITEGEOMETRIC
SERIES:
Considerthe
infinite
geometricseries
2
n-1
a
+ ar + ar + ... + ar + ...
then
a(1
-
r
n )
n-1
Sn =
a
+
ar
+
ar
+
L
+
ar
=
(r ≠
1)
2
1-
r
If
Sn →
S
as
n
→
∞,
then the series is
convergent and S is
itssum.
If
|r| < 1, then rn →
0 as n
→
∞
a(1
-
r
n )
∴
S
=
limSn = lim
1-
r
n→∞
n→∞
a
=
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MTH001
Elementary Mathematics
If
Sn increasesindefinitely
as n becomes verylarge then
the series is said to be
divergent.
EXERCISE:
Findthe
sum of the infinitegeometric
series:
93
2
+
+1+
+L
42
3
SOLUTION:
Here
we have
9
3/
2 2
a=
,
r=
=
4
9/4
3
Notethat
|r| < 1 So we canuse the
aboveformula.
a
∴
=
S
1-
r
9/4
=
1- 2 / 3
9
/ 4 9 3 27
=
=×=
1/
3 4 1 4
EXERCISE:
Find
a common fraction for
therecurring
decimal0.81
SOLUTION:
0.81=0.8181818181
...
=
0.81 + 0.0081 + 0.000081 +
...
which
is an infinite geometric
serieswith
0.0081
a
=
0.81,r =
=
0.01
0.81
a
∴
Sum =
1-
r
0.81
0.81
=
=
1
-
0.01
0.99
81
9
=
=
99
11
IMPORTANTSUMS:
n(n +
1)
n
1+ 2 +
3
+L
+
n
=
∑ k
=
1.
2
k
=1
n(n +
1)(2n +
1)
n
1
+
2
+
3
+L
+
n
=
∑ k
2 =
2
2
2
2
2.
6
k
=1
2
n2 (n
+
1)
⎡
n(n +
1)
⎤
n
1
+
2
+
3
+L
+
n
=
∑ k
=
=⎢
3
3
3
3
3
3.
⎣ 2 ⎥
4
⎦
k
=1
EXERCISE:
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MTH001
Elementary Mathematics
Sum
to n terms the series
1⋅5+5⋅11+9⋅17+...
SOLUTION:
Let
Tk denotethe
kth term of thegiven
series.
Then
Tk=
[1+(k-1)4]+ [0.5+(k-1)0.6]
=
(4k-3)+(0.6k-0.1)
=
4.6k-3.1
Now
Sk =
T1 + T2
+ T3 + ... +
Tn
n
∑T
=
k
k
=1
n
∑
(4.6k -
3.1)
=
k
=1
n
4.6∑
k
-3.1
=
k
=1
n(n +
1)
=
)
-
3.1
4.6(
2
=
2.3 n(n +
1)
-
3.1
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Table of Contents:
- Recommended Books:Set of Integers, SYMBOLIC REPRESENTATION
- Truth Tables for:DE MORGAN’S LAWS, TAUTOLOGY
- APPLYING LAWS OF LOGIC:TRANSLATING ENGLISH SENTENCES TO SYMBOLS
- BICONDITIONAL:LOGICAL EQUIVALENCE INVOLVING BICONDITIONAL
- BICONDITIONAL:ARGUMENT, VALID AND INVALID ARGUMENT
- BICONDITIONAL:TABULAR FORM, SUBSET, EQUAL SETS
- BICONDITIONAL:UNION, VENN DIAGRAM FOR UNION
- ORDERED PAIR:BINARY RELATION, BINARY RELATION
- REFLEXIVE RELATION:SYMMETRIC RELATION, TRANSITIVE RELATION
- REFLEXIVE RELATION:IRREFLEXIVE RELATION, ANTISYMMETRIC RELATION
- RELATIONS AND FUNCTIONS:FUNCTIONS AND NONFUNCTIONS
- INJECTIVE FUNCTION or ONE-TO-ONE FUNCTION:FUNCTION NOT ONTO
- SEQUENCE:ARITHMETIC SEQUENCE, GEOMETRIC SEQUENCE:
- SERIES:SUMMATION NOTATION, COMPUTING SUMMATIONS:
- Applications of Basic Mathematics Part 1:BASIC ARITHMETIC OPERATIONS
- Applications of Basic Mathematics Part 4:PERCENTAGE CHANGE
- Applications of Basic Mathematics Part 5:DECREASE IN RATE
- Applications of Basic Mathematics:NOTATIONS, ACCUMULATED VALUE
- Matrix and its dimension Types of matrix:TYPICAL APPLICATIONS
- MATRICES:Matrix Representation, ADDITION AND SUBTRACTION OF MATRICES
- RATIO AND PROPORTION MERCHANDISING:Punch recipe, PROPORTION
- WHAT IS STATISTICS?:CHARACTERISTICS OF THE SCIENCE OF STATISTICS
- WHAT IS STATISTICS?:COMPONENT BAR CHAR, MULTIPLE BAR CHART
- WHAT IS STATISTICS?:DESIRABLE PROPERTIES OF THE MODE, THE ARITHMETIC MEAN
- Median in Case of a Frequency Distribution of a Continuous Variable
- GEOMETRIC MEAN:HARMONIC MEAN, MID-QUARTILE RANGE
- GEOMETRIC MEAN:Number of Pupils, QUARTILE DEVIATION:
- GEOMETRIC MEAN:MEAN DEVIATION FOR GROUPED DATA
- COUNTING RULES:RULE OF PERMUTATION, RULE OF COMBINATION
- Definitions of Probability:MUTUALLY EXCLUSIVE EVENTS, Venn Diagram
- THE RELATIVE FREQUENCY DEFINITION OF PROBABILITY:ADDITION LAW
- THE RELATIVE FREQUENCY DEFINITION OF PROBABILITY:INDEPENDENT EVENTS